AN 


\-   ELEMENTARY   TREATISE 


6  m  MENSURATION 


^  rElLCTICAL   GEOMETRY 


I'JGIvTnER    Wt?H 


iOUS  PROBLEMS  OP  PMCIICAIj  lilPORTANCl 
MECHAT^IOS. 

WILLIAM  VOGDES,  LL.D, 

>r  5IATHBMAIIC8  XB  TH«  CSSTEAI.  HMH  SCnOOt  OT  j  . 
AIITHOR  OF  TH^  tlNITED  8TATK8  AEITfOHtlO. 

[PABT  FIESI.] 


PmLADEiafilA: 


E.  C\  &  J.  BIDDLE  &  CO.,  %  508  MINOR  ST. 

{Betwee*  Market  and  diMtfM,  and  J^^  and  Sixlh  BU.) 
1861. 


\^..'J^^^ 


J*> 


UCSB   LIBRARY 


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V 
V 


% 


Digitized  by  tine  Internet  Arcliive 

in  2007  witli  funding  from 

IVIicrosoft  Corporation 


littp://www.arcliive.org/details/elementarytreatiOOvogdiala 


AN 


ELEMENTARY   TREATitSJi: 


MENSURATION 


PRACTICAL   GEOMETRY, 


TOSETHKB   WITH 


NUMEROUS  PEOBLEMS  OF  PRACTICAL  IMPORTANCE 


MECHANICS. 


WILLIAM  VOGDES,  LL.D. 

LATE  PROFESSOR  OF  HATBEHATICS  IN  TDE  CENTRAL  HIGH  SCHOOL  OF  PHILA- 
DELPnlA,   AUTHOR  OP   THE   UNITED   STATES  ARITHMETIC. 


[PABT  FIEST.] 


PHILADELPHIA: 
E.  C.  &  J.  BIDDLE,  No.  508  MINOR  ST. 

(Between  Market  and  Chestnut,  and  Fifth  and  ,^xth  SU.) 

1862. 


Chamher  of  the  Controllers  of  Public  Sckoolt, ") 
First  School  District  of  Pennsylvania.  j 

Philadelphia,  November  15th,  1849. 
At  a  meeting  of  the  Controllers  of  Public  Schools,  First  Schooi 
District  of  Pennsylvania,  held  at  the  Controllers'  Chamber,  on  Tues- 
day, November  13th,  1849,  the  following  Resolution  was  adopted: — 
Resolved,  That  Vogdes'  Mensuration  be  introduced  as  a  Class  Book 
into  the  Grammar  Schools  of  the  District. 
From  the  Minutes. 

Robert  J   Hemphill,  Secretary. 


ExTEBED  according  to  act  of  Congress,  ia  the  year  1846,  by 

E.  C.  &  J.  BIDDLE, 

m  the  Qerk's  Office  of  the  District  Court  of  the  Eastern  District  of  Pennsyl- 


STEREOTTPED    »>    a..    JOHKSO*    &   CO. 
PHII.AnEI.PHIA. 

COLLIKS,   PKIKTBB 


PKEFACE. 


It  has  been  the  design  of  the  author,  in  the  following 
pages,  to  compile  a  work  adapted,  by  its  practical  cha- 
racter, to  the  wants  of  those  of  the  rising  generation, 
wtio,  not  being  able  to  command  a  collegiate  education, 
are  fitting  themselves  to  fill  useful  stations  in  society  as 
mechanics,  merchants,  &c.  With  tliis  end  in  view,  it 
was  deemed  inexpedient  to  devote  any  portion  of  the 
book  to  theoretical  demonstrations  of  the  principles  in- 
volved in  the  rules  given,  when  those  demonstrations  are 
based  upon  principles  illustrated  by  more  advanced 
branches  of  mathematical  science,  with  which  the  pupil 
is  supposed  to  be  unacquainted.  By  pursuing  this 
course,  room  was  afforded  for  the  introduction  of  more 
numerous  examples  illustrating  the  respective  rules,  than 
could  otherwise  have  been  given  without  the  enlarge- 
ment of  the  work  beyond  expedient  limits.  The  intro- 
duction of  these  examples,  it  is  believed,  wUl  enhance 
the  value  of  the  work  in  the  estimation  of  teachers  gene- 
rally, inasmuch  as  the  operation  of  the  rules  is  more 
likely  to  be  permanently  impressed  on  the  mind  of  the 
pupil,  by  long  continued  practice,  than  by  the  solution 
of  one  or  two  problems  only. 

In  a  treatise  on  Mensuration,  little  that  is  new  can  be 
looked  for,  other  than  the  collection  and  judicious 
arrangement  of  matters  not  heretofore  presented  to  the 
public  in  a  form  adapted  to  the  purposes  for  which  this 

3 


4  PREFACE. 

work  is  designed.  The  greater  portion  of  this  volume 
has  been  derived  from  the  works  of  Bonnycastle,  Has- 
weli,  Hutton,  Gregory,  and  Grier,  To  some  of  these, 
special  acknowledgment  of  the  obligation  has  been 
made  in  the  subsequent  pages. 

The  application  of  science  to  the  arts  of  industry  has 
oeen  one  of  tlie  most  potent  operative  causes  of  the 
rapid  increase  of  our  country  in  wealth  and  power.  In 
the  hope  that  this  compilation  may  be  found  a  useful 
assistant  to  the  teacher  who  is  engaged  in  preparing 
pupils  for  an  active  participation  in  these  industrial  pur- 
suits, the  author  submits  it  to  the  inspection  of  his  co- 
labourers  in  this  field. 

Philadelphia^  June  27,  1846. 


A  Key  to  this  work  has  been  published. 


CONTENTS. 


PRACTICAL  GEOMETRY. 

Pag« 

Definitions l& 

Instruments 23 

Geometrical  Problems 27 

MENSURATION  OF  SUPERFICIES. 

To  find  the  area  of  a  square 57 

To  find  the  side  of  a  square 58 

The  diagonal  of  a  square  being  given,  to  find  the  area 58 

The  area  of  a  square  being  given,  to  find  the  diagonal 59 

The  diagonal  of  a  square  being  given,  to  find  the  side 60 

To  cut  off  a  given  area  from  a  square  parallel  to  either  side.  60 

To  find  the  area  of  a  rectangle 61 

To  find  the  side  of  a  rectangle 62 

The  area  and  proportion  of  the  two  sides  of  a  rectangle  being 

given,  to  find  the  sides 63 

The  sides  of  a  rectangle  being  given,  to  cut  off  a  given  area 

parallel  to  either  side 64 

To  find  the  area  of  a  rhombus 65 

To  find  the  area  of  a  rhomboid 66 

To  find  the  side  of  a  rhombus  or  rhomboid 66 

To  find  the  area  of  a  triangle,  when  the  base  and  perpendi- 
cular are  given 67 

The  three  sides  of  a  triangle  being  given,  to  find  the  area. ...  68 
Any  two  sides  of  a  right-angled  triangle  being  given,  to  find 

the  third  side 69 

The  sum  of  the  hypothenuse  and  perpendicular,  and  the  base 
of  a  right-angled  triangle  being  given,  to  find  the  hypothe- 
nuse and  perpendicular 70 

To  determine  the  area  of  an  equilateral  triangle 71 

The  area  and  the  base  of  any  triangle  being  given,  to  find  the 

perpendicular  height 72 

The  proportion  of  the  three  sides  of  a  triangle  being  given,  to 

find  the  sides  of  a  triangle  corresponding  with  a  given  area  73 
The  base  and  perpendicular  of  any  triangle  being  given,  to 

find  the  side  of  an  inscribed  square 74 

The  three  sides  of  any  triangle  being  given,  to  find  the  length 

of  the  perpendicular  which  will  divide  it  into  two  right-  - 

angled  triangles "^H 

1*  5 


6  CONTENTS. 

Pag« 

The  area  and  base  of  a  triangle  being  given,  to  cut  off  a  given 
part  of  the  area  by  a  line  running  from  the  angle  opposite 

the  base 76 

The  area  and  base  of  a  triangle  being  given,  to  cut  off  a  tri- 
angle containing  a  given  area  by  a  line  running  parallel  to 

one  of  its  sides 77 

The  area  and  two  sides  of  a  triangle  being  given,  to  cut  off  a 
triangle  containing  a  given  area,  by  a  line  running  from  a 
given  point  in  one  of  the  given  sides,  and  falling  on  the 

other 78 

To  find  the  area  of  a  trapezium 79 

To  find  the  area  of  a  trapezoid 80 

To  find  the  area  of  a  regular  polygon 81 

To  find  the  area  of  a  regular  polygon  when  one  of  its  equal 

sides  only  is  given. . . .« 82 

When  the  area  of  any  regular  polygon  is  given  to  find  the 

side 83 

To  find  the  area  of  an  irregular  right-lined  figure  of  any  num- 
ber of  sides 83 

To  find  the  area  of  a  mixtilineal  figure,  or  one  formed  by 

right  lines  and  curves 85 

To  find  the  circumference  of  a  circle  when  the  diameter  is 
given,  or  the  diameter  when  the  circumference  is  given. . .     87 

To  find  the  area  of  a  circle 88 

The  area  of  a  circle  being  given,  to  find  the  diameter  or  cir- 
cumference      89 

To  find  the  area  of  a  circular  ring 89 

The  diameter  or  circumference  of  a  circle  being  given,  to  find 

the  side  of  an  equivaleiit  square 9C 

The  diameter  or  circumference  of  a  circle  being  given,  to  find 

the  side  of  the  inscribed  square 9i 

To  find  the  diameter  of  a  circle  equal  in  area  to  any  given 

superficies 92 

The  diameter  of  a  circle  being  given,  to  find  another  contain- 
ing a  proportionate  quantity 93 

To  find  the  length  of  any  arc  of  a  circle 93 

To  find  the  area  of  a  sector  of  a  circle 97 

To  find  the  area  of  a  segment  of  a  circle 100 

To  find  the  area  of  a  circular  zone 105 

To  find  the  area  of  a  lune 107 

To  find  the  area  of  a  part  of  a  ring,  or  of  the  segment  of  a  sector  1 09 

CONIC  SECTIONS. 

Definitions 110 

To  describe  an  ellipse 113 

The  ellipse  and  its  parts 114 

To  describe  a  parabola 120 

The  parabola  and  its  parts 120 

To  construct  the  hyperbola 124 

The  hyperbola  and  its  parts 124 


CONTENTS. 


MENSURATION  OF  SOLIDS. 

Page 
Definitions 130 

To  find  the  area  of  the  surface  of  a  cube. ...  133 

The  area  of  the  surface  of  a  cube  being  given,  to  find  the 
side 133 

To  find  the  solidity  of  a  cube 134 

To  find  the  side  of  a  cube  the  solidity  being  given 135 

To  find  the  solidity  of  a  parallelopipedon 136 

To  find  the  solidity  of  a  prism 136 

To  find  the  convex  surface  of  a  cylinder 137 

To  find  the  solidity  of  a  cylinder 138 

To  find  the  curve  surface  of  a  cylindric  ungula 139 

To  find  the  solidity  of  a  cylindric  ungula 139 

To  find  the  convex  surface  of  a  cylindric  ring 140 

To  find  the  solidity  of  a  cylindric  ring 141 

The  solidity  and  thickness  of  a  cylindric  ring  being  given,  to 

find  the  inner  diameter 143 

To  find  the  surface  of  a  right  cone  or  pyramid 142 

To  find  the  surface  of  the  frustum  of  a  right  cone  or  pyramid  144 

To  find  the  solidity  of  a  cone  or  pyramid 145 

To  find  the  solidity  of  the  frustum  of  a  cone  or  pyramid 146 

The  solidity  and  altitude  of  a  cone  being  given,  to  find  the 

diameter  of  the  base 148 

The  solidity  and  diameter  of  the  base  of  a  cone  being  given, 

to  find  the  altitude 149 

The  altitude  of  a  cone  or  pyramid  being  given,  to  divide  it 

into  two  or  more  equal  parts,  by  sections  parallel  to  the 

base,  to  find  the  perpendicular  height  of  each  part 149 

To  find  the  solidity  of  an  ungula,  when  the  section  passes 

through  the  opposite  extremities  of  the  ends  of  the  frustum  150 

To  find  the  solidity  of  a  cuneus  or  wedge 151 

To  find  the  solidity  of  a  pi'ismoid 152 

To  find  the  convex  surface  of  a  sphere 153 

To  find  the  solidity  of  a  sphere  or  globe 153 

The  convex  surface  of  a  globe  being  given,  to  find  its  diameter  154 

The  solidity  of  a  globe  being  given  to  find  the  diameter 155 

To  find  the  solidity  of  the  segment  of  a  sphere 155 

To  find  the  solidity  of  a  frustum  or  zone  of  a  sphere 156 

To  find  the  solidity  of  a  circular  spindle,  its  length  and  middle 

diameter  being  given 157 

To  find  the  solidity  of  the  frustum  of  a  circular  spindle,  its 

length,  the  middle  diameter,  and  that  of  either  of  the  ends. 

being  given 158 

To  find  the  solidity  of  a  spheroid,  its  two  axes  being  given  . .  159 
To  find  the  solidity  of  the  middle  frustum  of  a  spheroid,  its 

length,  the  middle  diameter,  and  that  of  either  of  the  ends 

being  given ICO 

To  find  the  solidity  of  the  segment  of  a  spheroid 102 

To  find  the  solidity  of  a  paraboloid '63 


8  CONTENTS. 

Vage 
To  find  the  solidity  of  the  frustum  of  a  parabolo  d,  when  its 

ends  are  perpendicular  to  the  axis  of  the  solid 164 

To  find  the  solidity  of  a  parabolic  spindle 166 

To  find  the  solidity  of  the  middle  frustum  of  a  parabolic  spindle  166 

To  find  the  solidity  of  a  hyperboloid 166 

To  find  the  solidity  of  the  frustum  of  a  hyperboloid 167 

REGULAR  BODIES. 

To  find  the  solidity  of  a  tetraedron 16 

To  find  the  side  of  a  tetraedron 170 

To  find  the  solidity  of  an  octaedron 170 

To  find  the  side  of  an  octaedron 171 

To  find  the  solidity  of  a  dodecaedron 1 71 

To  find  the  side  of  a  dodecaedron 172 

To  find  the  solidity  of  an  icosaedron 173 

To  find  the  side  of  an  icosaedron 173 

To  find  the  surface  of  any  regular  body 174 

To  find  the  solidity  of  any  regular  body ;.  1 76 

BALLS  AND  SHELLS. 

Weights  and  dimensions  of  balls  and  shells 176 

Piling  of  balls  and  shells 182 

CARPENTERS'  RULE 185 

The  use  of  the  Sliding  Rule 186 

TIMBER  MEASURE. 

To  find  the  content  of  a  board  or  piece  of  lumber  one  inch  in 

thickness 188 

To  find  the  content  of  any  piece  of  lumber,  whose  thickness 

is  more  than  an  inch 189 

HEWN  TIMBER. 

To  find  the  solid  content  of  squared  or  four-sided  timber 190 

The  breadth  of  a  board  being  given,  to  find  how  much  in 

length  will   make   a   square   foot,  or   any  other   required 

quantity 192 

When  the  board  is  wider  at  one  end  than  at  the  other,  and 

the  quantity  is  taken  from  the  less  end 192 

To  find  how  much  in  length  will  make  a  solid  foot  or  any 

other  required  quantity 193 

ROUND  TIMBER. 

To  find  the  solidity  of  round  or  unsquared  timber 194 

To  find  the  number  of  cord  feet  in  a  log 196 

To  find  the  number  of  solid  or  square  feet  that  a  piece  of 

round  timber  will  hew  to  when  squared 197 


CONTENTS.  a 

8AW-L0GS. 

Page 

To  find  the  number  of  feet,  board  measure,  that  a  log  contsiins  197 

MASONS' WORK 198 

BRICKLAYERS'  WORK 200 

CARPENTERS'  AND  JOINERS'  WORK 202 

SLATERS' AND  TILERS' WORK 203 

PLASTERERS'  WORK 204 

PAINTERS'  WORK 206 

GLAZIERS'  WORK 207 

PLUMBERS' WORK 208 

PAVERS'  WORK 209 

VAULTED  AND  ARCHED  ROOFS. 

To  find  the  solid  content  of  a  circular,  elliptical,  or  Gothic 

vaulted  roof 211 

To  find  the  concave  or  convex  surface  of  a  circular,  elliptical, 

or  Gothic  vaulted  roof 213 

To  find  the  solidity  of  a  dome 214 

To  find  the  superficial  content  of  a  dome 214 

To  find  the  solid  content  of  a  saloon 215 

To  find  the  superficies  of  a  saloon 216 

GAUGING. 

The  gauging  rule 217 

The  use  of  the  gauging  rule 219 

The  gauging  or  diagonal  rod 221 

To  find  the  content  of  a  cask  of  the  first  variety 222 

To  find  the  content  of  a  cask  of  the  second  variety 223 

To  find  the  content  of  a  cask  of  the  third  variety 224 

To  find  the  content  of  a  cask  of  the  fourth  variety 225 

To  find  the  content  of  a  cask  by  four  dimensions 226 

To  find  the  content  of  any  cask  from  three  dimensions 227 

To  ullage  a  standing  cask 228 

To  uMage  a  lying  cask 228 

To  find  the  ullage  by  the  gauging  rule 229 

To  find  the  content  of  a  cask  by  the  mean  diameter 229 

TONNAGE. 

Carpenters'  tonnage 231 

Government  tonnage 232 


10  CONTENT/S. 

CENTRE  OF  GRAVITY sfS 

FALLING  BODIES 240 

GRA,VITIES  OF  BODIES 245 

SPECIFIC  GRAVITY 246 

TABLE  OF  SPECIFIC  GRAVITIES 249 

THE  PENDULUM 253 

In  any  latitude  to  find  the  length  of  the  pendulum  which  beats 
seconds 256 

CENTRES  OF  PERCUSSION  AND  OSCILLATION 257 

CENTRE  OF  GYRATION 259 

PRESSURE  OF  THE  AIR 260 

DISTANCES,  AS   DETERMINED   BY  THE   VELOCITY 
OF  SOUND 261 

MECHANICAL  POWERS. 

The  Lever 263 

Wheel  and  Axle 265 

The  Pulley 267 

The  Inclined  Plane 270 

The  Wedge 271 

The  Screw 272 

WHEELS 274 

HYDRAULICS 277 

PUMPS. 279 

WATER-WHEELS 280 

MISCELLANEOUS  QUESTIONS 284 

TABLE  CfF  AREAS  OF  CIRCULAR  SEGMENTS 295 


TABLES. 


a 


1    £. 

1   ss 

£ 

< 

n    1 

1 

1-1  •^  o 

II   ^ 

mo 

II    ^ 

Ph  £-3 

1^ 

,-10000 

i-H  rj*  0  0 
0 

II 

"    00 

-<  -w  "-^  0  0  0 

0  OD  —  T"  0 

CO  Tj*  0*  00  0 

— <  rj«  ^ 

II                          1 

2- 

*--  «  x*  0  -^ 

(M  •^  0  10  00 
.-<  r}<  J— 

«§S§S8 
cTi     2^88 

II            ""i 

•»l>nTf  0  Tt*  Tj*  0  0  0 
woTj*  CI  Q  CO  0  Tf<  0 

0          w  c»  :c  t^  ao 
0  lO  0*  Tt< 

■-H   CO   ^ 

0 

11 


12  TABLES. 

Table  II. — ^Lineal  Measure. 


Inebos. 

Ganter'» 
Link. 

Feet. 

Yards. 

Fa- 
thoms. 

Rods, 
Polesjor 
Perches. 

Gunter's 
Chains. 

Fur- 
longs. 

Mile. 

m 

12 

36 
72 

198 

792 

7920 

63360 

=      1 

% 

100 

1000 

8000 

=       1 

3 

6 

16d 

66 

660 

5280 

=       1 

2 

H 

22 

220 

1760 

=       1 

21 

11 

110 

880 

=       1 

4 

40 

320 

=       1 

10 

80 

=       1 

8 

=     1 

Table  III. — Cubic  or  Solid  Measure. 


Cubic  Inches. 

Cubic  Feet. 

Cubic  Yards. 

Cub.  Poles, 
Rods,  or 
Perches. 

Cub. 
Fui» 
longs. 

612 

Cub. 
Mile. 

1728 

46656 

7762392 

496793088000 

2544358061056000 

27 

4482^ 

287496000 

147197952000 

=                 1 

166f 

10648000 

5451776000 

=             1 

64000 

32768000 

Table  IV. — Other  Measures. 


40  cubic  feet  of  round  timber  make 
50  cubic  feet  of  hewn  timber    " 
40  cubic  feet     .        -        -       " 
128  cubic  feet,  or  8  feet  in  length,  and  } 

4  in  breadth,  and  4  in  height,  make  3 
241  cubic  feet,  or  lOj  feet  in  length,    > 

I5  in  breadth  and  1  in  height  make  5 
282  cubic  inches  -        -        -         « 
231  cubic  inches 
277.274  cubic  inches  (Eng.) 
268^  cubic  inches 
2150.42  cubic  inches 


1  ton,  T. 

-    1  ton,  T. 

1  ton  of  shipping. 

1  cord  of  wood. 

1  perch  of  stone. 

1  gallon,  ale  measure. 

1  gallon,  wine  measure. 

1  imperial  gallon. 

1  gallon,  dry  measure. 

1  bushel 


EXPLANATION  OF  THE  CHAEACTERS 
USED  IN  THIS  WORK. 


+  denotes  plus,  or  more.  The  sign  of  addition,  signify- 
ing that  the  numbers  between  which  it  is  placed  are  to  be 
added  together.  Thus,  8  -f-  5,  denotes  that  5  is  to  be  added 
to  8.  Geometrical  lines  are  generally  represented  by  capi- 
tal letters.  Thus,  A  B  -f  C  D,  signifies  that  the  line  A  B  is 
to  be  added  to  the  line  C  D. 

—  denotes  minus,  or  less.  The  sign  of  subtraction,  sig- 
nifying that  the  latter  of  the  two  numbers  between  which  it 
is  placed  is  to  be  taken  from  the  former.  Thus,  4  —  2,  de- 
notes that  3  is  to  be  taken  from  4.  In  geometrical  lines, 
also,  A  B  —  CD,  signifies  that  the  line  C  D  is  to  be  sub- 
tracted from  the  line  A  B. 

X  denotes  into,  or  by.  The  sign  of  multiplication,  sig- 
nifying that  the  numbers  between  which  it  is  placed  are  to 
be  multiplied  together.  Thus,  7x5,  denotes  that  7  is  to 
be  multiplied  by  5.  In  geometrical  lines,  also,  A  B  x  C  D, 
signifies  that  the  number  of  units  in  the  line  A  B  is  to  be 
multipHed  by  the  number  of  units  in  the  fine  C  D.  Instead 
of  the  sign  Xj  a  point  is  sometimes  employed.  Thus, 
A  B .  C  D,  is  the  same  as  A  B  x  C  D. 

-T-  denotes  divided  by.  The  sign  of  division,  signifying 
that  the  former  of  the  two  numbers  between  which  it  is 
placed  is  to  be  divided  by  the  latter.  Thys,  6-7-3,  denotes 
that  6  is  to  be  divided  by  3.  This  is  also  expressed  by 
placing  the  dividend  above  a  line  and  the  divisor  below  it. 

Thus,  .-r  denotes  that  6  is  to  be  divided  by  3.     In  geometri- 
o 

cal  lines,  also,  A  B  -;-  C  D,  signifies  that  the  line  A  B  is  to 

A  B 
be  divided  by  the  line  C  D,  or  thus,  -p^rri. 

...  "I  are  proportionals,  signifying  that  the  numbers 
'"  an  '  l  ^6t^66n  which  they  are  placed  are  proportional. 
]  ■  r*Thus,  as  2  :  4  : :  8  :  16,  denotes  that  the  number 

J  2  has  the  same  proportion  to  4  that  8  has  to  1(». 


1 4  EXPLANATIONS. 

=  denotes  equal  to.  The  sign  of  equality,  signifying 
that  the  numbers  between  which  it  is  placed  are  equal  to 
each  other.  Thus,  2  poles  +  2  polos  =  4  poles  =  22 
yards  =  1  chain  =  100  links. 

(  )  the  parenthesis ;  [  ]  the  crotchet ;  and  {  }  the  brace, 
are  signs  mad§  use  of  to  connect  two  or  more  quantities 
together,  and  they  are  synonymous  with  regard  to  their 
application ;  for 

[(7  +  4)-5]x8  =  (ll-5)x8  =  6x8  =  48, 
is  the  same  as 

{ (7  +  4)  -  5}  X  8  =  ( 1 1  -  5)  X  8  =  6  X  8  =  48. 

The  parenthesis  which  includes  the  7  and  4,  serves  as  a 
chain  to  link  them  together,  and  shows  that  they  are  to  be 
added  together  before  the  number  5  is  subtracted ;  and  the 
brace  also  shows  that  the  numerals  which  it  includes  must 
be  operated  upon ;  and  the  result  multiplied  by  the  num- 
ber 8. 

*  This  sign  is  placed  above  a  quantity,  signifying  that  the 
quantity  is  to  be  squared. 

Thus,  (5  +  2)''  =  7^  =  7  X  7  =  49. 

^  This  sign  is  placed  above  a  quantity,  signifying  that  the 
quantity  is  to  be  cubed. 

Thus,  [(7  +  9)  —  8]3  =  (16  —  8)3  =  83  =  8  X  8  x8=512. 

v'  is  a  radical  sign,  signifymg  that  the  quantity  before 
which   it  is  placed  is  to  have  the  square  root  extracted. 

Thus, 

>/[(8  +  6)  -(4  X  2)y  +  {[(12  X  9)  -^  8]  -  [8  +  (9  -«)} 
+  {[(7  +  5)xl2]-[3x(4-2)]}  = 
^{(14-8)«+R108-3)-(8+7)]  +  [(12xl2)-(3x2)]} 
==  ^  [6^  +  (36  —  15)  +  (144  -^  6)]  =  >/  (36  +21  +  24) 
=  ^81  =  9. 

-^  is  a  radical  sign,  signifying  that  the  quantity  before 
which  it  is  placed  is  to  have  the  cube  root  extracted. 

Thus,  ^[(6  X  4  X  3)  —  8]  =  ^(72  -  8)  ==  ^64  =  4. 

.*.  denotes  therefore. 

JL  denotes  a  perpendicular. 

-<  denotes  an  angle. 

A  denotes  a  triangle. 


PRACTICAL  GEOMETRY. 


DEFINITIONS. 


§  1,  Practical  Geometry  is  a  mechanical  method  of  de- 
scribing mathematical  figures,  by  means  of  the  scale  and 
compasses,  or  other  instruments  proper  for  the  purpose.  It 
is  founded  upon  the  properties  and  relations  of  certain  mag- 
nitudes, which  may  be  found  treated  at  large  in  the  works 
of  Euclid  and  other  authors.  The  definitions  of  the  princi- 
pal figures  are  as  follows : 

1.  A  point,  considered  mathematically,  is  that  which  has 
no  parts  or  dimensions,  but  merely  position. 

2.  A  line  is  length  without  breadth,  and  its  bounds  or 
extremes  are  points. 

3.  A  right  or  straight  line  is  that  which  lies       

evenly  between  its  extreme  points  ;  as  A  B. 

4.  A  broken  line  is  one  which  changes      ^ ^ 

its   direction  at  intervals  so  large  that       /  \ 

they  can  be  perceived ;  as  A  B  C  D.       ^/  \p 

5.  A   curved   line  is  one  which 
changes  its  direction  at  intervals  so  ^ 
small  that  they  cannot  be  perceived  ;  as  A  B. 

6.  A  superficies  is  that  which  has  *'^ 
length  and  breadth  only ;  and  its  bounds 
or  extremes  are  fines  ;  as  A  B  C  D. 


B 

7.  A  plane,  or  plane  superficies,  is  that  Avhich  is  every- 
where perfectly  flat  and  even.  Or,  in  other  Avords,  it  is 
that  with  which  a  right  line  will  every  way  coincide. 


8.  A  body,  or  solid,  is  that  which   has  d 
length,  breadth,  and  thickness,  and  its  bounds, 
or  extremes,  are  superficies  ;  as  A  B  C  D. 

C 


15 


16 


PRACTICAL    GEOMETRY. 


0.  A  plane  rectilineal  angle  is  the 
inclination  or  opening  of  two  right 
linos,  which  meet  in  a  point  without 
cutting  each  other ;  as  A  B  C.     / 

Here  it  is  to  be  observed  that  the  greater  or  less  length 
of  the  lines  makes  no  alteration  in  the  angle. 

10.  One  right  line  is  said  to  be  perpen-  ^ 

dicular  to  another,  when  the  angles  on  each 
side  of  it  are  equal.    -^/ 

Thus  A  B  is  perpendicular  to  C  D.  An- 
gles are  of  three  kinds ;  being  either  right, 
acute,  or  obtuse. 


11.  A  right  angle  is  that  which  is  formed  by  two  right 
lines,  that  are  perpendicular  to  each  other ;  as    . 
ABC. 

Any  angle  differing  from  a  right  angle,  whether 
it  be  greater  or  less,  is  called  an  oblique  angle, 
and  the  lines  that  form  it  are  called  obhque  b 
lines. 


12.  An  acute  angle  is  that  which  is  less 
than  a  right  angle  ;  as  A  B  C. 


13.  An  obtuse  angle  is  that  which  is 
greater  than  a  right  angle,  as  ABC. 

14.  A  figure  is  a  space  bo\inded  by 
one  or  more  lines. 

15.  All  plane  figures  bounded  by 
three  right  lines  are  called  triangles, 
and  receive  different  denominations  according  to  the  nature 
of  their  sides  and  angles. 


16.  An     equilateral    triangle    is    that 
which  has  all  its  sides  equal ;  as  A  B  C. 


PRACTICAL    GEOMETRY. 


17 


17.  An  isosceles  triangle  is  that  which 
has  only  two  of  its  sides  equal ;  as  A  BC. 


18.  A  scalene  triangle  is  that  which 
has  all  its  three  sides  unequal ;  as 
ABC. 


10.  A  right-angled  triangle  is  that  which 
has  one  right  angle  ;  the  side  opposite  to 
the  right  angle  is  called  the  hypothenuse, 
and  the  other,  two  sides  the  legs  ;  as  A  B  C, 
where  A  B  is  the  hypothenuse,  and  B  C, 
A  C  the  two  legs,  or  base  and  perpendi- 
cular. 

Any  triangle  differing  from  a  right-angled  one  is  frequently 
called  an  oblique-angled  triangle. 


20.  An  obtuse-angled  triangle  is  that 
which  has  one  obtuse  angle ;  as  A  C  B, 
where  C  is  the  obtuse  angle. 


21.   An  acute-angled  tnangle  is  that  which 
has  all  its  angles  acute ;  as  A  B  C. 


22.  All  plane  figures  bounded  by  four  right  lines,  are 
•ailed  quadrangles,  or  quadrilaterals;  and  receive  different 
.;ames  according  to  the  nature  of  their  sides  and  angles. 


23.  A  square  is  a  quadrilateral,  whose 
•ides  are  all  equal,  and  its  angles  all  right 
angles  ;  as  A  B  C  D. 

2* 


18  PRACTICAL    GEOMETRY. 

A  square  is  also  an  instrument  used  by  artificers  for  what 
is  called  squaring  their  work ;  being  of  various  forms,  as  the 
T  square,  normal  square,  &c. 

A/ 

24.  A  rhombus  is  a  quadrilateral 
whose  sides  are  all  equal,  but  its  angles 
not  right  angles  ;  as  A  B  C  D. 


This  figure,  by  mechanics,  is  generally  called  a  lozenge ; 
and  both  it  and  the  square  belong  to  the  class  of  parallelo- 
grams. 


25.  A  parallelogram  is  a  quadri-  ^j 
lateral  whose  opposite  sides  are  / 
parallel ;  as  A  B  C  D.  /_ 


26.  A  rectangle  is  a  parallelogram  ^ 
whose  angles  are  all  right  angles ; 
as  A  B  C  D. 

27.  A  rhomboid  is  a  parallelogram  V 
whose  angles  are  not  right  angles ;  / 
as  A  B  C  D.  / 

B'— 


28.  A   trapezium    is    a    quadrilateral         -^ 
which  hath  not  its  opposite  sides  parallel ; 
as  A  B  C  D. 

B 


29.  A  trapezoid  is  a  quadrilateral,  having 
two  of  its  opposite  sides  parallel ;  as  A  B 
CD. 


30.  The  right  line  joining  any  two 
opposite  angles  of  a  quadrangle,  or  quad- 
rilateral, is  called  its  diagonal ;  as  B  D  in 
the  figure  A  B  C  D. 

B 


PRACTICAL    GICOMETRY. 


Id 


31.  All  plane  figures  contained  under  more -than  four 
sides  are  called  polygons  ;  and  receive  different  names,  ac- 
cording to  the  number  of  their  sides,  or  angles. 

32.  Thus,  polygons  having  five  sides  are  called  pentagons; 
those  of  six  sides,  hexagons,  those  of  seven,  heptagons ;  and 
so  on. 

iiS.  A  regular  polygon  is  that  which  has  all  its  sides  as 
well  as  its  angles  equal  to  each  other,  and  if  the  sides  or 
angles  are  unequal,  it  is  called  an  irregular  polygon. 

An  equilateral  triangle  is  also  a  regular  polygon  of  three 
sides,  and  a  square  is  one  of  four  sides. 

34.  Parallel  right  lines  are  such  as  are 

everjT\'here  at  an  equal  distance  from  each  a b 

other ;    or  which,  if  infinitely  produced,     

would  never  meet ;  thus  A  B  is  parallel 
to  CD. 

35.  The  base  of  any  figure  is  that  side  on  which  it  is  sup- 
posed to  stand  ;  and  its  altitude  is  the  perpendicular  falling 
upon  the  base  from  the  opposite  angle. 

36.  An  angle  is  usually  denoted 
by  three  letters,  the  one  which  stands 
at  the  angular  point  being  always  to 

be  read  in  the  middle,  as  A  B  C,  C  a 

B  D,  D  B  E,  &c. 

37.  A  circle  is  a  plane  figure  bounded 
by  a  curve  line  called  the  circumference  or 
periphery,  which  is  everywhere  equidis- 
tant from  a  point  within,  called  its  centre  ; 
and  is  formed  by  the  revolution  of  a  right 
line  (O  A)  about  one  of  its  extremities  (O), 
which  remains  fixed. 

88.  The  centre  of  a  circle  is  the  point  (O)  about  which  it 
is  described ;  and  the  circumference  or  periphery  is  the  iine 
or  boundary  A  B  C  A,  by  which  it  is  contained. 

The  circumference  itself,  as  well  as  the  space  which  is 
bounded  by  it,  is  also,  for  the  sake  of  conciseness,  sometimes 
called  a  circle. 


20 


PRACTICAL  GEOMETRY. 


39.  The  radius  of  a  circle  is  a  right  line 
drawn  from  the  centre  to  the  circumference ; 
as  0  A. 

A  tangent  is  a  line  touching  a  circle,  and 
which  produced,  does  not  cut  it,  as  B  A  C. 


40.  The  diameter  of  a  circle  is  a  right 
line  passing  through  the  centre,  and 
terminated  both  ways  by  the  circumfe- 
rence ;  as  A  B. 


41.  An  arc  of  a  circle  is  any  part  of  its 
circumference,  or  periphery  ;  as  A  B. 


42.  A  chord  is  a  right  line  which  joins 
the  extremities  of  an  arc ;  as  A  B. 


43.  A  segment  of  a  circle  is  the  space 
contained  between  an  arc  and  its  chord ; 
as  A  B  C. 


44.  A  sector  is  the  space  contained  be- 
tween an  arc  and  the  two  radii  drawn  to 
its  extremities  ;  as  A  B  C. 


PRACTICAL    GEOMETRY. 


21 


45.  A  zone  is  a  part  of  a  circle  included  ^ 
between  two  parallel  chords  and  their  inter- 
cepted arcs  ;  as  A  B  C  D.  ^ 


46.  The  versed  sine  or  height  of  an  arc, 
is  that  part  of  the  diameter  contained  be-  ^ 
tween  the  middle  of  the  chord  and  the  arc  ; 
as  D  B,  or  D  E. 


47.  A  lune  is  the  space  included  between 
the  intersecting  arcs  of  two  eccentric  circles ; 
as  A  B  C.  Ak-  ^B 

48.  A  semicircle  is  a  half  of  a  circle ;  a  quadrant  is  a 
quarter  of  a  circle ;  a  sextant  the  sixth  part  of  it,  and  an 
octant  the  eighth  part,  where  it  may  be  observed  that  these 
names  are  often  applied  to  instruments  used  for  taking 
angles. 

49.  The  circumference  of  every  circle  is  supposed  to  be 
divided  into  360  equal  parts  called  degrees ;  each  degree 
into  60  equal  parts,  called  minutes;  and  each  minute  into  60 
equal  parts,  called  seconds. 

50.  The  measure  of  any  right -lined  a 

angle  is  an  arc  of  a  circle  contained  be-        r 

tween  the  two  lines  which  form  that      ' 

angle,   the   angular   point    being    the     \ 

centre ;  thus  the  angle  A  O  B  is  mea-      \  / 

sured  by  the  arc  m  n.  **-...-'* 

The  angle  is  estimated  bj^  the  number  of  degrees,  minutes, 
&c.  contained  in  the  arc ;  whence  a  right  angle  is  an  angle 
of  90  degrees  or  |  of  the  circumference. 


AXIOMS. 


1.  An  axiom  is  an  established  principle,  or  self-evident 
truth,  requiring  no  other  conviction  than  that  which  arises 
from  a  proper  understanding  of  the  terms  in  which  if  is  pro- 
posed. 


9,2  PRACTICAL    GEOMETRY. 

2.  Things  which  are  equal  to  the  same  thing  are  equal  to 
'Sach  other. 

3.  If  equals  be  added  to  equals  the  wholes  will  be  equal. 

4.  If  equals  be  taken  from  equals  the  remainders  will  be 
equal. 

5.  If  equals  be  added  to  unequals  the  wholes  will  be  un- 
equal. 

6.  If  equals  be  taken  from  unequals  the  remamders  will 
be  unequal. 

7.  Things  which  are  half,  double,  or  any  number  of  times 
the  same  thing,  are  equal. 

8.  The  whole  is  greater  than  its  part 

9.  Every  whole  is  equal  to  all  its  parts  taken  together. 

10.  All  right  angles  are  equal  to  each  other. 

11.  Angles  that  have  equal  measures,  or  arcs,  are  equal. 

12.  Things  which  coincide,  or  fill  the  same  space,  are 
identical,  or  mutually  equal  in  all  their  parts. 

13.  Two  straight  lines  cannot  enclose  a  space. 


REMARKS. 

1    A  problem  is  something  proposed  to  be  done. 

2.  The  perimeter  of  a  figure  is  the  sum  of  all  its  sides 
taken  together. 

3.  The  sum  of  any  two  sides  of  a  triangle  is  greater  than 
the  third  side. 

4.  In  any  triangle  the  sum  of  the  three  angles  is  equal  to 
two  right  angles. 

T).  Every  triangle  is  half  the  parallelogram  which  has  the 
same  base  and  the  same  altitude. 

6.  An  angle  inscribed  in  a  semicircle  is  a  right  angle. 

7.  All  angles  in  the  same  segment  of  a  circle  are  equal  to 
each  other. 


PRACTICAL    GEOMMPRY.  23 

8.  Triangles  that  have  all  the  three  angles  of  the  one 
respectively  equal  to  all  the  three  angles  of  the  other,  aro 
called  equiangular  triangles,  or  similar  triangles. 

9.  In  similar  triangles  the  like  sides,  or  sides  opposite  to 
the  equal  angles,  are  proportional. 

10.  The  areas  or  spaces  of  similar  triangles  are  to  each 
other  as  the  squares  of  their  like  sides. 

1 1 .  The  areas  of  circles  are  to  each  other  as  the  squares 
of  their  diameters,  radii,  or  circumference. 

12.  Similar  figures  are  such  as  have  the  same  number  of 
sides,  and  the  angles  contained  by  their  sides  respectively 
equal. 

13.  The  areas  of  similar  figures  are  to  each  as  the  squares 
of  their  like  sides. 

14.  If  three  quantities  are  proportional,  the  middle  one  is  re- 
peated, and  the  first  is  to  the  second  as  the  second  is  to  the  third. 

In  such  a  case  the  middle  quantity  is  a  mean  proportional 
between  the  other  two,  and  the  last  is  a  third  proportional  to 
the  first  and  second;  but,  if  there  are  four  proportional  quan- 
tities, the  last  is  called  a  fourth  proportional  to  the  other  three 

INSTRUMENTS. 

§2,  The  principal  instruments  used  in  describing  or  con- 
structing geometrical  figures  are  as  follows  : 

THE    DIVIDERS    OR   COMPASSES. 


The  plain  compasses  consist  of  two  inflexible  rods  of  brass, 
revolving  upon  an  axis  at  the  vertex,  and  furnished  with 
steel  points. 


24 


PRiMfriCAL    GEOMETRY. 


THE    PARALLEL   RULER. 


The  parallel  ruler  consists  of  two  flat  pieces  of  ebony,  con- 
nected together  with  brass  bars,  having  their  extremities 
equidistant,  by  which  contrivance,  when  the  ruler  is  opened, 
the  sides  necessarily  move  in  parallel  lines. 


THE    SCALE    OF    EQUAL    PARTS. 


UsM 


-n-f 


±d 


The  scale  of  equal  parts  consists  of  a  certain  number  of 
equal  portions  of  any  convenient  length,  the  extreme  one  or 
the  left  hand  being  subdivided  into  ten  equal  parts,  and 
is  called  the  unit  of  the  scale,  and  the  rest  being  numbered 
1,2,  3,  &c. 

In  most  scales  an  inch  is  taken  for  a  common  measure,  and 
what  an  inch  is  divided  into  is  generally  set  at  the  end  of  the 
scale. 

This  scale  is  used  in  laying  down  any  distance,  as  inches, 
feet,  chains,  miles,  &c.  The  several  divisions  may  be  con- 
sidered as  feet,  for  example,  the  decimal  subdivision  would 
be  tenths  of  a  foot.  So  also  each  of  the  principal  divisions 
may  be  regarded  as  ten  inches,  ten  feet,  &c.,  and  in  this 
case  the  decimal  subdivision  will  represent  inches,  feet,  &c., 
respectively.  This  scale  is  limited  to  two  figures,  or  any 
number  less  than  100  may  be  readily  taken  ;  but  if  the 
number  should  consist  of  three  places  of  figures,  the  value 
01  the  third  figure  cannot  be  exactly  ascertained,  and  m  this 
case  it  is  better  to  use  a  diagonal  scale,  by  which  any  num- 
ber consisting  of  three  places  of  figures  may  be  exactly 
tound. 

Let  it  be  required  to  take  from  the  scale  a  line  equal  to 
five  inches  and  eight-tenths. 


PRACTICAL    GKOMtTRY. 


Place  one  foot  of  the  dividers  at  5  on  the  right,  and  ex- 
tend the  other  to. 8,  which  makes  the  eighth  of  the  small 
divisions.  The  dividers  will  then  embrace  the  required 
distance. 


THE    DIAGONAL   SCALE    OF    EQUAL    PARTS. 


«  \b  g] 


The  construction  of  this  scale  is  as  follows  : 
Having  prepared  a  ruler  of  convenient  breadth  for  your 
scale,  draw  near  the  edges  thereof  two  right  lines,  a  e,  c  f, 
parallel  to  each  other;  divide  one  of  these  hnes,  as  a  e,  into 
equal  parts,  according  to  the  size  of  your  scale,  and  through 
each  of  these  divisions  draw  right  lines  perpendicular  to 
a  e,  to  meet  cf;  then  divide  the  breadth  into  ten  equal  parts, 
and  through  each  of  these  divisions  draw  right  lines  parallel 
to  a  e  and  c/;  divide  the  lines  a  b,  c  d,  into  ten  equal  parts, 
and  from-  the  point  a,  to  the  first  division  in  the  line  c  d, 
draw  a  diagonal  line  ;  then  parallel  to  that  line,  draw  diago- 
nal lines  through  all  the  other  divisions,  and  the  scale  is  com- 
plete. Then,  if  any  number  consisting  of  three  places  of 
figures,  as  408, 1^  required  from  the  larger  ^ca\e,fd,  you 
must  place  one  foot  of  the  compasses  on  the  figure  4,  on  the 
liney  (/,  then  the  extent  from  4  to  the  point  d  will  represent 
400.  The  second  figure  being  6,  count  six  of  the  smaller 
divisions  from  d  towards  c,  and  the  extent  from  4  to  that 
point  will  be  460.  Mo'-e  both  points  of  the  compasses  down- 
wards till  they  are  on  the  eighth  parallel  line  below _/"  d,  and 
open  them  a  little  till  the  one  point  rests  on  the  vertical  line 
drawn  through  4,  and  the  other  on  the  diagonal  line  drawn 
through  6  ;  the  extent,  then,  in  the  compasses  will  represent 
468.  "  In  the  same  manner  the  quantities  46.8,  4.68,  0.468. 
&c.,  are  measured. 

There  are  generally  two  diagonal  scales  laid  down  on  the 
same  face  of  the  instrument,  the  unit  of  the  one  being  double 
that  of  the  other,  and  commencing  on  opposite  ends  of  the 
scale. 


PRACTICAL   GEOMETRY. 


THE    SCALE    OF    CHORDS. 

c 

3o/        "^N^/             \ 

1  •                ii 

,(/ 

B 

Draw  the  lines  B  A,  B  C,  at  right  angles  to  each  other, 
and  with  any  convenient  distance,  B  A,  describe  the  arc 
A  s  C  ;  divide  it  into  90  equal  parts,  and  join  A  C.  From 
A  as  a  centre,  with  the  distances  A  10,  A  20,  &c.,  describe 
the  arcs  10,  10,  20,  20,  &c.,  meeting  the  line  A  C.  Fill 
up  the  separate  degrees,  which  are  not  marked  in  the  diagram 
to  prevent  confusion,  and  the  scale  is  complete.  It  is  evi- 
dent, by  inspection,  that  the  chord  of  60  is  equal  to  the  radius, 
as  shown  by  the  letter  r  upon  the  rule  ;  which  distance  ia 
therefore  always  to  be  taken  in  laying  down  anglea. 


THE    SEMICIRCULAR   PROTRACTOR. 


The  protractor  is  a  semicircular  piece  of  brass  divided  into 
180  degrees,  and  numbered  each  way  from  end  to  end ;  that 
is,  from  A  to  B,  and  from  B  to  A.     There  is  a  small  notch 


PRACTICAL    GEOMETRY.  27 

in  the  middle  of  the  diameter  A  B,  denoting  the  centre  of 
the  protractor.  In  some  boxes  of  mathematical  instruments 
this  is  omitted,  and  the  degrees  are  transferred  to  the  border 
of  the  plain  scale. 


GUNTER  S  SCALE. 

Gunter's  scale,  commonly  of  two  feet  in  length,  contains 
on  one  side  the  lines  of  ihe  plain  scale,  already  described, 
and  on  the  other  corresponding  logarithmic  lines. 


PROBLEM  I. 

§  3.  To  describe  from  a  given  centre  the  circumference 
of  a  circle  having  a  given  radius. 

Let  A  be  the  given  centre,  and  A  B 
the  given  radius. 

Place  one  foot  of  the  dividers  at  A,  and 
extend  the  other  leg  until  it  shall  reach  b[ 
toB. 

Then  turn  the  dividers  around  the  leg 
at  A,  and  the  other  leg  will  describe  the 
required  circumference. 


PROBLEM  n. 

Through  a  given  point  A,  to  draw  a  line  parallel  to  a  given 
line,  D  E. 

Lay  the  edge  of  the  parallel  ruler    ^ a ^ 

upon  D  E,  and  move  it  upwards  till  it 
reaches  the  point  A,  through  which 
draw  B  C,  and  it  will  be  parallel  to  D  E.     ° 


PRACTICAL  GEOMETRY. 


PROBLEM  in. 


To  lay  off  on  a  given  line,  as  AB,  a  distance  equal 
to  CD. 

f 

BXAMPLE. 

Let  C  D  be  the  distance  to  be  laid  off,     c d 

and  A  B  the  given  line.  ^ f    ^ 

Place  one  foot  of  the  dividers  at  C,  and  extend  the  other 
leg  until  the  foot  reaches  D. 

Then,  raising  the  dividers,  place  one  foot  at  A,  and  mark 
with  the  other  the  distance  AE,  this  will  evidently  be 
equal  to  C  D. 

PROBLEM  IV. 

To  lay  do^vn  a  line  of  given  length,  on  a  scale  of  a  given 
number  to  the  inch,  to  determine  how  many  parts  of  it  are 
to  be  represented  on  the  paper  by  a  distance  equal  to  the 
unit  of  the  scale. 

EXAMPLE. 

If  a  line  320  feet  in  length  is  to  be  laid  down  on  paper, 
on  a  scale  of  40  feet  to  the  inch,  what  length  must  be  taken 
from  the  scale  ? 

Divide  the  length  of  the  line  by  the  number  of  parts  which 
is  represented  by  the  unit  of  the  scale  ;  the  quotient  wiA 
give  the  number  of  parts  which  is  to  be  taken  from  the 
scale. 

Here  320-5-40=8  the  number  of  parts  to  be  taken  from 
the  scale. 

PROBLEM  V. 

The  length  of  the  line  being  given  on  the  paper,  to  deter 
mine  the  true  length  of  the  line  which  it  represents. 

EXAMPLE.  V 

The  length  of  the  line  on  the  paper  is  4.75  inches,  and 
the  scale  is  one  of  20  feet  to  the  inch  ;  what  is  the  true  leiiyli 
of  the  line  ? 


PRACTICAL    GEOMETRY.  29 

Take  the  line  in  your  dividers  and  apply  it  to  the  scale, 
ftnd  note  the  number  of  units  and  parts  of  a  unit  to  which 
It  is  equal ;  then  multiply  this  number  by  the  number  of 
parts  which  the  unit  of  the  scale  represents,  and  the  product 
will  be  the  length  of  the  line. 

Here  4.75x20=95  feet,  the  length  of  the  line. 


PROBLEM  VI. 

To  make  an  angle  of  any  proposed  number  of  degrees. 
c 


1.  Draw  any  line  A  B,  and  having  taken  first  60  degrees 
from  the  scale  of  chords,  describe  with  this  radius  the  arc  n  m, 

2.  Take  in  like  manner  the  chord  of  the  proposed  num- 
ber of  degrees  from  the  same  scale,  and  apply  it  from  n  to  m. 

3.  Then  if  the  line  A  C  be  drawn  from  the  point  A 
through  m,  the  angL  BAG  will  be  that  required. 

Angles  greater  than  90  degrees  are  usually  made  by  first 
laying  off  90  degrees  upon  the  arc  n  in,  and  then  the  remain- 
ing part. 

This  problem  may  be  performed  by  the  protractor. 

Place  the  central  notch  of  the  instrument  upon  A,  and  the 
edge  along  A  B  ;  make  a  point  in  against  the  proposed  num- 
ber of  degrees,  and  through  it  draw  the  line  A  C 


3* 


30 


PRACTICAL    GEOMETRY. 


PROBLEM  VIL 

Any  angle  BAG  being  given,  to  find  the  number  of  de- 
grees it  contains. 


?V--. 


1.  From  the  angular  point  A,  with  the  chord  of  60 
degrees,  describe  the  arc  n  m,  cutting  the  lines  A  B,  A  C  in 
the  points  n  and  m. 

2.  Then  take  the  distance  n  m,  and  apply  it  to  the  scale 
of  chords,  and  it  will  show  the  degrees  required. 

And  if  the  distance  n  m  be  greater  than  90°,  it  must  be 
taken  at  twice,  and  each  part  applied  separately  to  the  scale. 
This  problem  may  be  performed  by  the  protractor. 

Place  the  central  notch  of  the  instrument  upon  A,  and 
the  edge  along  A  C,  and  observe  the  number  of  degrees  cut 
by  the  line  A  B,  which  will  show  the  degrees  required. 


PROBLEM  VIIL 


To  divide  a  given  line  AB  into  two  equal  parts. 


1.  From  the  points  A  and  B,  as 
centres,  with  any  distance  greater 
than  half  A  B,  describe  arcs  cutting 
each  other  in  n  and  m. 

2.  Through  these  points,  draw 
tne  line  n  c  in,  and  the  point  c,  where 
it  cuts  A  B,  will  be  the  middle  of  the 
'ine  required. 


'm 


PRACTICAL    GEOMETRY. 


31 


PROBLEM  IX. 


To  divide  a  given  angle  ABC  into  two  equal  parts. 


1.  From  the  point  B,  with  any  radius, 
dpscribe  the  arc  A  C  ;  and  from  A  and  C, 
with  the  same,  or  any  other  radius,  describe 
arcs  cutting  each  other  in  n. 

2.  Then,  through  the  point  n  draw  the 
line  B n,  and  it  will  bisect  the  angle  ABC, 
as  was  required. 


PROBLEM  X. 

From  a  given  point  C,  in  a  given  right  line  A  B,  to  erect 
a  perpendicular. 

Case  I.  When  the  point  is  near  the  middle  of  the  line. 

1.  On  each  side  of  the  point  C 
take  any  two  equal  distances  Cn, 
Cm. 

2.  From  n  and  m,  with  any  ra- 
dius greater  than  n  C  or  m  C,  de- 
scribe arcs  cutting  each  other  in  s. 

3.  Then  through   the   point   8,  ^ 
draw  the  line  sC,  and  it  will  be         n 
the  perpendicular  required. 

Case  II.  When  the  point  is  at,  or  near,  the  end  of  the 
line. 

Supposing  C  to  be  the  given  point,  as  before. 

1.  Take  any  point  o,  and  with  the 
radius  or  distance,  oC,  describe  the 
arc  mCn,  cutting  A  B  in  m  and  C. 

2.  Through  the  centre  o,  and  the 
point  m,  draw  the  line  m  o  n,  cutting 
the  arc  m  C  n  in  n. 

3.  Then  from  the  point  n,  draw 

the  line  nC,  and  it  will  be  the  per-  ^ } 

pendicular  required. 


> 

n 

/ 

/ 

3/ 

4 

/ 

/ 

/ 

92  PRACTICAL   GEOMETRY. 

• 

The  same  otherwise. 

1.  Set  one  leg  of  the  compasses 
on  B,  and  with  any  extent  Bw 
describe  an  arc  mp  ;  then  set  oflf 
the  same  extent  from  m  to  n. 

2.  Then  join  m  n,  and  from  n 
as  a  centre  with  the  extent  m  n  as 
radius,  describe  an  arc  g ;  pro- 
duce mn  to  q,  and  the  Une  join- 
ing q  B  will  be  perpendicular  to 
AB.  ^ 

Another  method. 

1.  From  any  scale  of  equal  parts 
take  a  distance  equal  to  3  divisions, 
and  set  it  from  B  to  m. 

2.  And  from  the  points  B  and 
m,  with  the  distances  4  and  5, 
taken  from  the  same  scale,  describe 
arcs  cutting  each  other  in  n. 

3.  Through  the  points  n,  B,  draw 
the  line  BC,  and   it  wiU  be   the     ^  m     .s     a 
perpendicular  required. 

The  same  thing  may  also  be  readily  done,  by  an  instru- 
ment in  the  form  of  a  square,  or  by  the  plain  scale. 

PROBLEM  XL 

From  a  given  point  C,  without  a  given  line  A  B,  to  let 
fall  a  perpendicular. 

Case  I.  When  the  point  is  nearly  opposite  to  the  middle 
of  the  line. 

1.  From  the  point  C,  with  any 
radius,  describe  the  arc  n  m,  cut- 
ting A  B  in  n  and  m. 

2.  From  the  points  n,  m,  with 

the  same  or  any  other  radius,  de-  a         ^ sLjb 

scribe  two  arcs  cutting  each  other 
in  8. 

3.  Through  the  points  C,  s, 
draw  the  Hne  C  G  a,  and  C  G  will 
be  the  peroendicular  required 


PRACTICAL   GEOMETRY. 


33 


v- 


'-^ ^- 


<'B 


B 


This  problem,  also,  may  be  performed  like  the  last  by 
means  of  a  square. 

Case  II.  When  the  point  is  opposite  or  nearly  opposite 
to  the  end  of  the  line. 

1.  Take  any  point  7»,  in  the 

hne  A  B,  and  from  C  draw  the  '^^c 

line  C  m. 

2.  Bisect  the  hne  C  m,  or  di- 
vide it  into  two  equal  parts  in  the 
point  n. 

3.  From  n,  with  the  radius  nm, 
or  nC,  describe  the  arc  CDm, 
cutting  AB  in  D. 

4.  Then  through  the  point  C, 

draw  the  hne  C  D,  and  it  will  be  the  perpendicular  required. 
This  method  may  also  be  used  in  the  first  case,  if  the  line 
A  B,  when  necessary,  be  produced. 
The  same  otherwise. 

1.  From  A,  or  any  other  point  in 
A  B,  with  the  radius  A  C,  describe  the 
arcs  C,  D. 

2.  And  from  any  other  point  n,  in 
AB,  with  the  radius  nC,  describe 
another  arc  cutting  the  former  in  C,  D. 

3.  Then  through  the  points  C,  D, 
draw  the  line  C  E  D,  and  C  E  will  be 
the  perpendicular  required. 

Perpendiculars  may  be  more  easily 
raised,  and  let  fall,  in  practice,  by  means 
of  a  square,  or  other  instrument  proper 
for  this  purpose. 


-c.o 


I  £ 


---^S- 


PROBLEM  XII. 


To  draw  a  perpendicular,  from  any  angle  of  a  triangle 
A  B  C,  to  its  opposite  side. 


1.  Bisect  either  of  the  sides  con- 
taining the  angle  from  which  the  per- 
pendicular is  to  be  drawn,  as  B  C  in 
the  point  n 


r>--. 


34 


PKACTICAL   GEOMETRY. 


2.  Then  with  the  radius  nC, 
and  from  the  centre  n  describe  an 
arc  cutting  AB,  (or  AB  produced  if 
necessary,  as  in  the  second  figure,) 
in  the  point  D ;  the  line  joining  G 
D  will  be  perpendicular  to  A  B  or 
4  B  produced. 


PROBLEM  XIII. 

To  trisect,  or  divide  a  right  angle  ABC,  into  three  equal 
parts. 

1.  From  the  point  B,  with  any 
radius  B  A,  describe  the  arc  A  C,  cut- 
ting the  legs  B  A,  BC,  in  A,  C. 

2.  From  the  point  A,  with  the 
radius  A  B,  or  B  C,  cross  the  arc  A  C 
in  n ;  and  with  the  same  radius,  from 
the  point  C,  cross  it  in  m. 

3.  Then  through  the  points  m,  n, 
draw  the  lines  Bm,  Bn,  and  they 
will  trisect  the  angle  as  was  required. 

By  this  means  the  circumference  of  any  given  circle  may 
be  divided  into  12  equal  parts  ;  and  thence  by  bisection  into 
24,  48,  «&c. 


PROBLEM  XIV. 


At  a  given  point  D,  to  make  an  angle  equal  to  a  given 
angle  ABC. 


B  m  AD  r  E 

1.  From  the  point  B,  with  any  radius,  describe  the  arc  n 
m,  cutting  the  lines  B  A,  B  C,  in  the  points  w,  n. 


PRACTICAL    GEOMEIRY.  35 

2.  Draw  the  line  D  E,  and  from  the  point  D,  with  the 
same  radius  as  before,  describe  the  arc  rs. 

3.  Take  the  distance  m  n,  on  the  former  arc,  and  apply  it 
to  the  arc  r  s,  from  r  to  s. 

4.  Then  through  the  points  D,  s,  draw  the  line  D  F,  and 
the  angle  E  D  F  will  be  equal  to  the  angle  A  B  C,  as  was 
required. 


PROBLEM  XV. 

To  draw  a  line  parallel  to  a  given  hne  A  B. 

C^E  I.  When  the  parallel  line  is  to  pass  through  a  given 
point  C. 

1.  Take  any  point  m  in 
the  line  A  B,  and  from  the 
point  C  draw  the  hne  C  ni. 

2.  From    the    point   m, 
with   the  radius   mC,  de- 
scribe the  arc  C  n,  cutting 
A  B  in  n ;  and  with  the  same  radius,  from  the  point  C,  de- 
scribe the  arc  m  r. 

3.  Take  the  distance  C  n,  and  apply  it  to  the  arc  m  r,  from 
m  to  r;  then  through  the  points  C,  r,  draw  the  line  D  C  r  E, 
and  it  will  be  parallel  to  A  B,  as  was  required. 

Case  11.  When  the  parallel  hne  is  to  be  at  a  given  dis 
tance  from  A  B. 

1 .  From  any  two  points  r,  s, 
in  the  line  A  B,  with  a  radius 
equal  to  the  given  distance, 
describe  the  arcs,  n,  m. 

2.  Then  draw  the  hne  C  D 
to   touch   these   arcs  without   ■'^        »"  » 
cutting  them,  atd  it  will  be  parallel  to  A  B,  as  was  re<|uired 


c 

f 

' 

/    ^^ 

/ 

/                '^■s. 

/ 

/ 

'                                                                          X' 

D          n 

m        Q 

1 

0                                                                                           -^ 

36  PRACTICAL    GEOMETRY. 


PROBLEM  XVI. 

To  divide  a  given  line  A  B  into  any  proposed  number  of 
equal  parts. 

1.  From  one  end  of  the  line  A,  6  ** 
draw  A  m,  making*  any  angle  with  ^<.••-'C'^ 
AB;  and  from  the  other  end  B,  ^  *?--'' \''  N  \'i 
draw  B  n,  making  an  equal  angle  j^  ^-'-r''  \  \  \  '«> 
A  B  n.                                                    ;\     \     \    \,''X" 

2.  In  each  of  the  lines  A  wi,  B  n,      \  \     \ ,,- V'* 
beginning  at  A  and  B,  set  off  as      V--'5'''^ 
many  equal  parts,  of  any  conve-      '' 

nient  length,  as  A  B  is  to  be  divided  into.  , 

3.  Then  join  the  points  A  5,  1,  4,  2,  3,  &c.,  and  A  B  will 
be  divided  as  was  required. 

B  n  may  be  drawn  parallel  to  A  m,  by  means  of  the 
parallel  ruler,  or  the  same  may  be  done  by  taking  the  aic 
A  n  equal  to  B  m. 

Another  method. 

1.  Through  the  point  B  draw  the 
indefinite  line  C  E,  making  an  angle 
with  A  B. 

2.  Take  any  point  E  in  that  line, 
through  which  draw  E  D  parallel  to 
B  A,  and  set  off  as  many  equal  parts 
E  H,  H  G,  &c.,  from  E  toward  D,  as  d 
<he  line  A  B  is  to  be  divided  into. 

3.  Through  the  points  D,  A,  draw  the  line  DA,  producing 
it  till  it  meets  E  C  in  C  ;  then  lines  drawn  from  C  through 
ihe  points  F,  G,  H,  will  divide  the  Hne  A  B,  into  the  re- 
ouired  number  of  parts. 


PRACTICAL    GEOMETRY. 


37 


PROBLEM  XVII. 

To  find  the  centre  of  a  given  circle,  or  of  one  already 
described. 

1.  Draw  any  chord,  A  B,  and  bisect 
it  with  the  perpendicular,  C  D. 

2.  Bisect  C  D,  in  like  manner,  with 
the  chord  E  F,  and  the  intersection,  O, 
Will  be  the  centre  required ;  observing 
that  the  bisection  of  the  chords,  and  the 
raising  of  the  perpendicular,  may  be  per- 
xormed  as  in  Problems  VIII.  and  X. 


PROBLEM  XVIIL 

To  describe  the  circumference  of  a  circle  through  any 
three  given  points,  A,  B,  C,  provided  they  are  not  in  a  right 
line. 

1.  From  the  middle  point  B,  draw  the 
lines,  or  chords,  B  A,  B  C  ;  and  bisect 
them  perpendicularly  with  the  lines 
meeting  each  other  in  O. 

2.  Then  from  the  point  of  intersec- 
tion, O,  with  the  distance  O  A,  O  B,  or 
O  C,  describe  the  circle  ABC,  and  it 
will  be  that  required. 


PROBLEM  XEX. 


To  draw  a  tangent  to  a  given  circle,  that  shall  pass  through 
a  given  point  A. 

Case  I. — When  the  point  A  is  in  the  circumference  of 
the  circle. 

M ,A 

1 .  From  the  given  point  A,  to  the 
centre  of  the  circle,  draw  the  radius 
AO. 

2.  Then  through  the  point  A,  draw 
B  C  perpendicular  to  O  A,  and  it  will 
be  the  tangent  required. 


38 


PRACTICAL    GEOMETRY. 


Case  II. —  When  the  given  point,  A,  is  without  the  circle. 

1.  To  the  given  point  A,  from  the 
centre  O,  draw  the  line  O  A,  and  bisect 
it  in  n. 

2.  From  the  point  n,  with  the  radius 
71  A,  or  n  O,  describe  the  semicircle 
ABO,  cutting  the  given  circle  in  B. 

3.  Then  through  the  points  A,B»  draw 
the  line  A  B,  and  it  will  be  the  tangent 
required. 


PROBLEM  XX. 
To  two  given  right  lines,  A,B,  to  find  a  third  proportional. 


1.  From  any  point,  C,  draw  two  right  lines,  making  any 
angle,  F  C  G. 

2.  In  these  lines  take  C  E  equal  to  the  first  term  A,  and 
C  G,  C  D,  each  equal  to  the  second  term  B. 

3.  Join  E  D,  and  draw  G  F  parallel  to  it ;  and  C  F  will 

be  the  third  proportional  required. 

That  is  C  E  (A) :  C  G  (B) :  :C  D  (B) :  C  F. 


PROBLEM  XXL 


To  three  given  right  lines,  A,B,C,  to  find  a  fourth  pro- 
portional. 


E  O 


PRACTICAL    GEOMETRY.  39 

1 .  From  any  point,  D,  draw  two  right  lines,  making  any 
angle  G  D  H. 

2.  In  these  lines  take  D  F  equal  to  the  first  term  A,  D  E 
equal  to  the  second  B,  and  D  H  equal  to  the  third  C 

3.  Join  F  E,  and  draw  H  G  parallel  to  it,  then  D  G  will  be 
the  fourth  proportional  required. 

That  is  D  F  (A) :  D  E  (B) : :  D  H  (C) :  D  G. 

PROBLEM  XXII. 

Between  two  given  right  lines  A,  B,  to  find  a  mean  pro- 
portional. 


1.  Draw  any  right  Une,  in  which  take  C  E,  equal  to  A, 
and  E  D,  equal  to  B. 

2.  Bisect  C  D  in  O,  and  with  O  D  or  O  C,  as  radius,  de- 
scribe the  semicircle  C  F  D. 

3.  From  the  point  E  draw  E  F  perpendicular  to  C  D,  and 
it  will  be  the  mean  proportional  required. 

That  is C  E  (A) :  E  F : :  E  F  :  ED  (B). 

PROBLEM  XXIIL 

To  divide  a  given  line,  A  B,  in  the  same  proportion  that 
another  given  line,  C,  is  divided. 

1.  From  the  point  A  draw  A  D 
equal  to  the  given  line  C,  and  making 
any  angle  with  A  B. 

2.  To  A  D  apply  the  several  divi- 
sions of  C,  and  join  D,B. 

3.  Draw  the  lines  4,4,  3,3,  &c., 
each  parallel  to  D  B,  and  the  line 
A  B  will  be  divided  as  was  required. 


40 


PRACTICAL   GEOMETRY. 


That  is,  the  parts  A,  1 ;  1,  2 ;  2,  3 ;  3,  4 ;  4,  B;  on  the 
line  A  B,  will  be  proportional  to  the  parts  0,  1  ;  1,  2 ;  2,  3 : 
3|  4 ;  4,  5 ;  on  the  line  C. 


PROBLEM  XXIV. 


Two  angles  of  a  triangle  being  given  to  find  the  third. 

1.  Draw  the  indefinite 
line  ABC;  at  any  point, 
as  B,  make  the  angle 
A  B  D  equal  to  one  of 
the  given  angles,  and  the 
angle  D  B  E,  equal  to 
the  other. 

2.  The  remaining  an- 
gle E  B  C  will  be  the 

third  angle  required,  because  these  three  angles  are  together 
«;qual  to  two  right  angles. 


PROBLEM  XXV. 

Two  sides  of  a  triangle,  and  the  angle  which  they  coiiiain, 
being  given  to  construct  the  triangle. 

1.  Let  A  and  B  be 
equal  to  the  given 
sides,  and  C  the  given 
angle. 

2.  Draw  the  indefi- 
nite line  D  E  ;  at  the 
point  D,  make  the 
angle  E  D  F  equal  to  »  a 
the  given  angle  C  ; 
then   take  D   G=A,  p 

D   H=B,  and  draw 

G  H ;  then  D  G  H  will  be  the  triangle  required. 


PRACTICAL    GEOMETRY. 


41 


PROBLEM  XXVI. 
A  side  and  two  angles  being  given  to  constnict  the  triangle. 

1.  The  two  angles  will  either  be 
adjacent  to  the  given  side,  or  the  one 
adjacent  and  the  other  opposite  ;  in 
the  latter  case,  find  the  third  angle  by- 
subtracting  the  sum  of  the  two  given 
angles  from  180  degrees ;  then  the  two 
adjacent  angles  will  be  known. 

2.  Draw  the  line  A  B  equal  to  the  given  side  ;  at  the 
point  A,  make  an  angle,  BAD,  equal  to  one  of  the  adjacent 
angles,  and  at  B,  an  angle  equal  to  the  other  ;  the  two  lines 
A  D,  B  E,  will  cut  each  other  at  C  ;  and  ABC  will  be  the 
triangle  required. 


PROBLEM  XXVn. 

Upon  a  given  right  line,  A  B,  to  make  an  equilateral  tri- 
anofle. 


1.  From  the  points  A  and  B,  with 
a  radius  equal  to  A  B,  describe  arcs 
cutting  each  other  in  C. 

2.  Draw  the  lines  A  C,  B  C,  and 
the  figure  A  C  B  will  be  the  triangle 
required. 

An  isosceles  triangle  may  be  formed 
in  the  same  manner,  by  taking  any 
distance  for  radius. 


--.  O--- 


42 


PRACTICAL   GEOMETRY. 


PROBLEM  XXVIIL 

The  three  sides  of  a  triangle  being  given,  to  describe  the 
triangle. 

Let  A,  B,  and  C,  be  the  sides. 

1.  Draw  a  line,  D  E,  equal  to  one  of  the  given  lines,  C. 

2.  From  the  point  D,  with  a  radius  equal  to  B,  describe 
an  arc. 


3.  And  from  the  point  E,  with  a  radius  equal  to  A,  de> 
scribe  another  arc,  cutting  the  former  in  F. 

4.  Then  draw  the  lines  D  F,  E  F,  and  D  F  E  will  be  the 
triangle  required. 


PROBLEM  XXIX. 


Upon  a  given  line,  A  B,  to  describe  a  square. 


D; 


1 .  From  the  point  B,  draw  B  C  per- 
pendicular, and  equal  to  A  B. 

2.  From  the  points  A  and  C,  with  the 
radius  A  B  or  C  B,  describe  two  arcs 
cutting  each  other  in  D ;  then  draw  the 
lines  A  D,  C  D,  and  the  figure  A  BC  D 
will  be  the  square  required. 

A  rhombus  may  be  made  on  the  given         ^  ^ 

line  A  B,  in  exactly  the  same  manner,  if  B  C  be  drav,'n  with 
tne  proper  obliquity,  instead  of  perpendicularly. 


PRACTICAL    GKOMKTRY. 


4A 


Another  method. 

1.  From  the  points  A,  B,  as  centres, 
with  the  radius  A  B,  describe  arcs 
crossing  each  other  in  m. 

2.  Bisect  A  m  in  n  ;  and  from  the 
centre  m,  with  the  radius  m  n,  cross 
the  other  two  arcs  in  C  and  D. 

3.  Then  draw  the  lines  A  D,  B  C, 
and  C  D,  and  A  B  C  D  will  be  the 
square  required. 


PROBLEM  XXX. 


To  describe  a  rectangle,  whose  length  and  breadth  shall 
be  equal  to  two  given  lines,  A  B,  and  C. 

1.  At  the  point  B,  in  the  given 
line  A  B,  erect  the  perpendicular 
B  D,  and  make  it  equal  to  C. 

2.  From  the  points  D,  A,  with 
the  radii  A  B  and  C  respectively, 
describe  two  arcs  cutting  each  other 
in  E  ;  then  join  E  A  and  E  D,  and 
A  B  D  E  will  be  the  rectangle  re- 
quired. 

A  parallelogram  may  be  descnbed   in  nearly  the  same 
manner. 


PROBLEM  XXXL 


In  a  given  triangle,  A  B  C,  to  inscribe  a  circle. 


1 .  Bisect  any  two  of  the  angles, 
as  A  and  B,  with  the  lines  A  O, 
and  B  O. 

2.  Then  from  the  point  of  in- 
tersection, O,  let  fall  the  perpen- 
dicular O  n,  upon  either  of  the 
sides,  and  it  will  be  the  radius  of 
the  circle  required. 


44  PRACTICAL    GEOMETRY. 


PROBLEM  XXXn. 


In  a  given  circle  to  inscribe  an  equilateral  triangle,  a  hexar 
gon,  or  a  dodecagon. 


FOR    THE    HEXAGON. 

1.  From  any  point,  A,  in  the  circum- 
ference, as  a  centre,  with  a  distance 
equal  to  the  radius  A  O,  describe  the 
arc,  B  O  F. 

2.  Join  the  pomts  A,  B,  or  A,  F,  and 
either  of  these  lines  being  carried  six 
times  round  the  circle  will  form  the 
hexagon  required. 

That  is,  the  radius  of  the  circle  is  equal  to  the  side  of  the 
inscribed  hexagon  ;  and  the  sides  of  the  hexagon  divide  the 
circumference  of  the  circle  into  six  equal  parts,  each  contain- 
ing 60  degrees. 


FOR   THE   EQUILATERAL   TRIANGLE. 

1.  From  the  point  A,  to  the  second  and  fourth  divisions, 
or  angles  of  the  hexagon,  draw  the  lines  A  C,  A  E. 

2.  Then  join  the  points  C,  E,  and  ACE  will  be  the  equi- 
lateral triangle  required  ;  the  arcs,  ABC,  C  D E, and E F  A, 
being  each  one-third  of  the  circumference,  or  120  degrees. 


FOR   THE    DODECAGON. 

Bisect  the  arc  A  B,  subtending  the  side  of  the  hexagon  in 
»fhe  point  n,  and  the  line  A  n  being  carried  twelve  times 
lound  the  circumference,  will  form  the  dodecagon  required ; 
<he  arc  A  n  being  30  degrees. 

Then,  if  A  n  be  again  bisected,  a  polygon  may  be  formed 
of  24  sides ;  and  by  another  bisection  a  polygon  of  48  sides ; 
and  so  on. 


PRACTICAL   GEOMETKV. 


45 


PROBLEM  XXXm. 
To  inscribe  a  square  or  an  octagon,  in  a  given  circle. 

FOR  TKB   SQUARE. 


1 .  Draw  the  diameters  B  D  and  A  C, 
intersecting  each  other  at  right  angles. 

2.  Then  join  the  points  A,B,  B,C, 
C,  A,  and  D,  A;  and  A  B  C  D  wiU  be 
ihe  square  required. 


FOR   THE    OCTAGON. 

Bisect  the  arc  A  B,  subtending  the  side  of  the  square,  in 
the  point  E,  and  the  line  A  E  being  carried  eight  times  round 
the  circumference  will  form  the  octagon. 

Also,  if  the  arc  A  E  be  again  bisected,  a  polygon  may  be 
formed  of  16  sides  :  and  by  another  bisection,  a  polygon  of 
33  sides  ;  and  so  on. 


PROBLEM  XXXIV. 
To  inscribe  a  pentagon,  or  decagon,  in  a  given  circle. 


FOR   THE    PENTAGON. 

1.  Draw  the  diameters  A  p,  n  m, 
at  right  angles  to  each  other,  and 
bisect  the  radius  O  n  in  r. 

2.  From  the  point  r,  with  the  dis- 
tance r  A,  describe  the  arc  A  s,  and 
from  the  point  A,  with  the  distance 
A  *,  describe  the  arc  s  B. 

3.  Join  the  points  A,  B,  and  the 
line  A  B  being  carried  five  times 
round  the  circle,  will  form  the  pentagon  required. 


46 


PRACTICAL    GEOMETRY. 
FOR  THE    DECAGON. 


Bisect  the  arc  A  E,  subtending  the  side  of  the  pentagon  in 
c,  and  the  line  A  c  being  carried  ten  times  round  the  cir- 
cumference will  form  the  decagon  required. 

Also,  if  the  arc  A  C  be  again  bisected,  a  polygon  of  20 
sides  may  be  formed  ;  and  by  another  bisection  a  polygon 
of  40  sides ;  and  so  on. 


PROBLEM  XXXV. 

In  a  given  circle,  to  inscribe  a  regular  heptagon. 

1.  From  any  point  A  in  the  circum- 
terence,  with  the  radius  A  O  of  the 
circle,  describe  the  arc  B  O  C,  cutting 
the  circumference  in  B  and  C 

2.  Draw  the  chord  B  C,  cutting  O  A 
m  D,  and  B  D,  or  C  D,  carried  seven 
times  round  the  circle  from  A,  will  form 
the  heptagon  required. 

An  exact  heptagon,  as  is  well  known,  cannot  be  inscribed 
in  a  circle  from  pure  geometrical  principles  ;  but,  the  above 
method  of  constructmg  it,  which  is  extremely  simple,  will 
be  found  sufficiently  accurate  to  answer  most  practical 
purposes ;  the  approximation  for  the  side  being  true,  as  far 
as  the  second  place  of  decimals,  inclusively. 


PROBLEM  XXXVL 

In  a  given  circle,  to  inscribe  any  regular  polygon. 

1.  Draw  the  diameter  A  B,  which 
divide  into  as  many  equal  parts  as  the 
ligure  has  sides. 

2.  From  the  points  A,  B,  as  centres, 
with  the  radius  A  B,  describe  arcs 
crossing  each  other  in  C. 

3.  From  the  point  C,  through  the 
second  division  of  the  diameter,  draw 
the  line  C  D ;  then  if  the  points  A  and 
D  be  joined,  the  line  A  D  will  be  the 
side  of  the  polygon  required. 


PRACTICAL    GEOMETRY. 


47 


It  may  be  observed,  that  in  the  construction  here  given. 
A  D  is  the  side  of  a  pentagon. 

This  rule  is  not  exact,  except  for  the  equilateral  triangle 
and  hexagon  ;  being  for  polygons  in  general  only  a  near 
approximation. 


PROBLEM  XXXVII. 

On  a  given  line  A  G,  to  describe  a  regular  polygon  of  any 
proposed  number  of  sides. 

1.  From  the  point  A,  w^ith 
the  distance  A  G,  describe  the 
semicircle  G  B  H,  which  di^nde 
into  as  many  equal  parts  as  H  a, 
a  B,  Be,  &c.,  as  the  polygon 
is  to  have  sides. 

2.  From  A  to  the  second  part 
of  the  division  draw  A  B,  and 
through  the  other  points  c,  d,  e, 
f,  &c.,  draw  the  lines  A  C,  A  D, 
A  E,  A  F,  &c. 

3.  Apply  the  distance  A  G  from  G  to  F,  from  F  to  E, 
from  E  to  I>,  from  D  to  C,  &c.,  and  join  B  C,  CD,  D  E, 
E  F,  &c.,  and  A  B  C  D  E  F  G  will  be  the  polygon  required. 


PROBLEM  XXXVIIL 
About  a  given  triangle  A  B  C,  to  circumscribe  a  circle. 

1.  Bisect  the  two  sides  A  B,  BC, 
with  the  perpendiculars  m  O,  and  n  O. 

2.  From  the  point  of  intersection  O, 
with  the  distance  OA,  OB,  or  OC,  de- 
scribe the  circle  A  C  B,  and  it  will  be 
that  required. 

If  any  two  of  the  angles  be  bisected, 
instead  of  the  sides,  the  intersection  of 
these  lines  will  give,  as  before  shown, 
the  centre  of  the  inscribed  circle. 


48 


PRACTICAL   GEOMETRY. 


PROBLEM  XXXIX. 

About  a  given  square,  A  B  C  D,  to  circumscribe  a  circle. 

1.  Draw  the  two  diagonals  AC  and 
B  D  intersecting  each  other  in  O. 

2.  Then  from  the  point  O,  with  the 
distance  O  A,  O  B,  O  C  or  O  D,  describe 
the  circle  A  B  C  D,  and  it  will  be  that 
required. 


PROBLEM  XL. 


To  circumscribe  a  square  about  a  given  circle. 


1.  Draw  any  two  diameters  no  and 
r  m  at  right  angles  to  each  other. 

2.  Then  through  the  points  m,  o,  r,  n, 
draw  the  lines  AB,  BC,  CD,  and 
D  A,  perpendicular  to  r  m,  and  n  o,  and 
A.  B  C  D  will  be  the  square  required. 


T 


PROBLEM  XLL 

About  a  given  circle  to  circumscribe  a  pentagon. 

1 .  Inscribe  a  pentagon  in  the  circle ; 
or  which  is  the  same  thing,  find  the 
pomts  m,n,  v,  r,s,  as  in Prob. XXXIV. 

2.  From  the  centre  o,  to  each  of 
these  points,  draw  the  radii  o  7i,  o  m, 
0  V,  0  r,  and  o  s. 

3.  Through  the  points  n,  m,  draw 
the  lines  A  B,  B  C,  perpendicular  to 
on,  om;  producing  them  till  they 
meet  each  other  at  B. 

Draw  in  Hke  manner  the  lines  C  D,  D  E,  E  A,  perpen- 
dicular to  0  V,  0  r,  and  o  «,  and  A  B  C  D  £  will  be  the 
pentagon  required. 


PRACTICAL    GEOMETRY. 


49 


Any  other  polygon  may  be  made  to  circumscribe  a  circle, 
by  first  inscribing  a  similar  one,  and  then  drawing  tangents 
to  the  circle  at  the  angular  points. 


PROBLEM  XLH. 
On  a  given  line,  A  B,  to  make  a  regular  pentagon. 

1.  Make  B  wi  perpendicular  to  A  B, 
and  equal  to  one  half  of  it. 

2.  Draw  A  m,  and  produce  it  till 
the  part  m  n  is  equal  to  B  m. 

3.  From  A  and  B  as  centres,  with 
the  radius  B  n,  describe  arcs  cutting 
each  other  in  o. 

4.  Then  from  the  point  o,  with  the 
same  radius,  or  with  o  A,  or  o  B,  de- 
scribe the  circle,  ABODE;  and  the  line  A  B,  appUed  five 
times  round  the  circumference  of  this  circle,  will  form  the 
pentagon  required. 

If  tangents  be  drawn  through  the  angular  points  A,  B,  C, 
D,  E,  a  pentagon  circumscribing  the  circle  will  be  formed ; 
and  if  the  arcs  be  bisected,  a  circumscribing  decagon  may  be 
formed ;  and  so  on. 


4"^ 


7^ 


Another  method. 


,    D 


t.  Produce  A  B  towards  n,  and 
at  the  point  B  make  the  perpendi- 
cular B  m,  equal  to  A  B. 

2.  Bisect  A  B  in  r,  and  from  r 
as  a  centre,  with  the  radius  r  m,  de- 
scribe the  arc  m  n,  cutting  the  pro- 
duced line  A  B  in  n. 

3.  From  the  points  A  and  B, 
with  the  radius  A  n,  describe  arcs  cutting  each  other  in  D ; 
and  from  the  points  A,  D,  and  B,  with  the  radius  A  B,  de- 
scribe arcs  cutting  each  other  in  C  and  E. 

4.  Then  if  the  lines  B  C,C  D,  D  E,  and  E  A,  be  drawn, 
ABODE  will  be  the  pentagon  required. 


50 


PRACTICAL    GEOMETRY. 


PROBLEM  XLHI. 
On  a  given  line,  A  B,  to  make  a  regular  hexagon. 


1.  From  the  points  A,  and  B,  as  cen- 
tres, with  the  radius  A  B,  describe  arcs 
cutting  each  other  in  0  ;  and  from  the 
point  O,  with  the  distance  O  A,  or  O  B, 
describe  the  circle  A  B  C  D  E  F. 

2.  Then  if  the  line  A  B  be  applied 
six  limes  round  the  circumference,  it 
will  form  the  hexagon  required. 


^,-- 


PROBLEM  XLIV. 
On  a  given  line,  A  B,  to  form  a  regular  octagon. 

1.  On  the  extremities  of  the  given 
line,  A  B,  erect  the  indefinite  perpen- 
diculars A  F,  and  B  E. 

2.  Produce  A  B  both  ways  to  m  and 
n,  and  bisect  the  angles  m  A  F  and 
n  B  E  with  the  hnes  A  H  and  B  C. 

3.  Make  A  H  and  B  C  each  equal  to 
AB,  and  draw  H  G,C  D,  parallel  to  AF,  m'- 
or  B  E,  and  also  each  equal  to  A  B. 

4.  From  G,  and  D,  as  centres,  with  a  radius  equal  to 
describe  arcs  crossing  A  F,  B  E,  in  F  and  E  ;  then  if 
FE,  and  E  D  be  drawn,  ABCDEFGH  will  be  the 
gon  required. 


AB, 
GF, 

octa- 


PROBLEM  XLV. 
To  make  a  figure  similar  to  a  given  figure,  ABODE. 

1 .  Take  A  6,  equal  to  one  of  the  sides 
of  the  figure  required,  and  from  the  angle 
A  draw  the  diagonals  A  c,  A  d.  « 

2.  From  the  points  b,  c,  d,  draw  b  c,cd,  ^ 
d  e,  parallel  to  B  C,  C  D,  D  E,  and  <" 
Ab  c  d  e  will  be  similar  to  A  B  C  D  E. 

The  same  thing  may  also  be  done  by   a 
making  the  angles  b,  c,  d,  e,  respectively 
equal  to  the  angles  B,  C,  D,  E. 


PRACTICAL    GEOMETRY. 
PROBLEM  XL VI. 


51 


To  make  a  triangle  equal  to  a  given  trapezium,  A  B  C  D. 

1.  Draw  the  diagonal  D  B,  and 
make  C  E  parallel  to  it,  meeting  the 
side  A  B  produced  in  E. 

2.  Join  the  points  D,  E,  and 
A  D  E  will  be  the  triangle  re- 
quired. 


PROBLEM  XL VII. 

To    make  a  triangle    equal   to   any   right-Iinod   figure, 
ABCDEFGA. 

1 .  Produce  the  side  A  B  both 
ways  at  pleasure. 

2.  Draw  the  diagonals  E  A, 
E  B;  and  by  the  last  problem, 
make  the  triangles  A  E  I,  B  E  K, 
equal  to  the  figures  A  E  F  G,  and 
B  E  D  C. 

3.  Draw  I  L,  K  M,  parallel  to 
E  A,  E  B. 

4.  Then  if  the  points  E,  L,  and  E,  M,  be  joined,  ELM 
will  be  the  triangle  required. 

And  in  the  same  manner  may  any  right-lined  figure  what- 
ever be  reduced  to  a  triangle. 


PROBLEM  XLVm. 

To  describe  a  square  that  shall  be  any  multiple  of  a  given 
square,  A  B  C  D. 

1.  Draw  the  diagonal  B  D,  and  on 
A  B,  A  D,  produced,  take  AG,  A  E, 
each  equal  to  B.D,  then  A  F,  the  square 
on  A  E,  will  be  double  the  square  A  C. 

2.  Draw  in  like  manner  the  diagonal 
B  G,  and  make  A  L,  A  H,  each  equal  to 
B  G ;  then  A  K,  the  square  on  A  H,  will 
be  triple  the  square  A  C. 


62 


PRACTICAL    GEOMETRY. 


3.  Proceed  in  the  same  manner  with  the  diagonal  B  L ; 
and  a  square  will  be  formed  which  is  quadruple  the  square 
A  C ;  and  so  on. 


PROBLEM  XLIX. 


To  inscribe  a  square  in  a  given  triangle,  ABC. 

1.  Through  the  ver- 
tex A,  parallel  to  B  C, 
draw  the  straight  line 
A  E,  and  from  C  raise 
a  perpendicular  to 
meet  it  in  D. 

2.  Make  DE  equal 
to  D  C  ;  join  E  B,  cutting  D  C  in  F. 

3.  Through  F  draw  F  G,  parallel  to  B  C,  and  through  H 
and  G,  draw  H  L,  G  K,  paraUel  to  D  C.  Then  L  H  G  K 
will  be  the  square  required. 


PROBLEM  L. 


In  a  given  circle  to  inscribe  a  polygon  of  any  proposed 
number  of  sides. 


1.  Divide360°by  the  numberof  sides 
of  the  figure  and  make  an  angle  A  O  B, 
at  the  centre,  whose  measure  shall  be 
equal  to  the  degrees  in  the  quotient. 

2.  Then  join  the  points  A,  B,  and 
apply  the  chord  A  B  to  the  circumfe- 
rence the  given  number  of  times,  and  it 
will  form  the  polygon  required. 


PROBLEM  LL 


On  a  given  line  A  B  to  form  a  regular  polygon  of  any 
proposed  number  of  sides. 

1,  Divide  360°  by  the  number  of  sides  of  the  figure,  and 
subtract  the  quotient  from  180  degrees. 


PRACTICAL    GEOMETRY. 


53 


2.  Make  the  angles  ABO  and  BAO 

each  equal  to  half  the  difference  last 
found ;  and  from  the  point  of  intersec- 
tion O,  with  the  distance  O A  or  OB, 
describe  a  circle. 

3.  Then  apply  the  chord  A  B  to  the 
circumference  the  proposed  number  of 
times,  and  it  will  form  the  polygon 
required. 


PROBLEM  LIT. 
To  describe  a  circle  within  or  without  a  regular  polygon. 


1.  Bisect   any  two  angles,  A  ED, 
EDC,  by  the   lines   EG,    DG,  and 
from  G  let  fall  G  F,  perpendicular  to    ^^ 
the  side  E  D. 

2.  Then  with  the  radius  GE  de- 
scribe the  outer  circle,  and  with  the 
radius  G  F  describe  the  inner  circle. 


PROBLEM  Llir. 

To  make  a  square  that  shall  be  nearly  equal  to  a  given 
circle. 

1.  In  the  given  circle  A  C  B  D,  draw 
the  two  diameters  A B,  CD,  cutting 
each  other  perpendicularly  in  the 
centre  O. 

2.  Bisect  the  radius  O  C,  in  E,  and 
trough  the  points  A,  E,  draw  the  chord 
AEF,  which  will  be  the  side  of  a 
square  that  is  nearly  equal  in  area  to 
ihe  circle. 

If  F  G  be  drawn  parallel  to  C  D,  it  will  be  nearly  equal 
to  I  of  the  circumference  of  the  circle. 


R* 


64 


PRACTICAL    GEOMETRY. 


PROBLEM  LIV. 

To  find  a  right  line  that  shall  be  nearly  equal  to  any  given 
arc  A  D  B  of  a  circle. 


1.  Divide  the  chord  AB  into  four 
equal  parts,  and  set  one  of  the  parts  A  C, 
on  the  arc  from  B  to  D. 

2.  Draw  C  D,  and  the  double  of  this 
line  will  be  nearly  equal  to  the  arc  A  D  B. 

If  a  right  line  be  made  equal  to  3| 
times  the  diameter  of  a  circle,  it  will 
be  nearly  equal  to  the  circumference. 


PROBLEM  LV. 

To  divide  a  given  circle  into  any  proposed  number  of 
parts,  that  shall  be  mutually  equal  to  each  other,  both  in  area 
and  perimeter. 

1.  Divide  the  diameter  A  B  into  the 
proposed  number  of  equal  parts,  at  the 
points  a,  b,  c,  d. 

2.  On  Aa,  Ab,  Ac,  Ad,  &c.,  as 
diameters,  describe  semicircles  on  one 
side  of  the  diameter  AB;  and  on  Bd, 
Be,  B6,  Ba,  &c.,  describe  semicircles 
on  the  other  side  of  the  diameter. 

3.  Then  the  corresponding  semicircles,  joinmg  each  other 
as  above,  will  divide  the  circle  in  the  manner  proposed. 


PROBLEM  LVL 

In  a  given  circle,  to  describe  three  equal  circles  which 
shall  touch  one  another,  and  also  the  circumference  of  the 
given  circle. 

From  the  centre  O,  let  the  right- 
lines  O  H,  O  D.  and  O  E  be  drawn, 
dividing  the  circumference  into 
three  equal  parts  in  the  points  H, 
D,  and  E  ;  join  D  E,  and  in  O  E 
produced  take  E  F,  equiU  to  on^ 


PRACTICAL    GF.OMLTRY 


55 


half  of  D  E ;  draw  D  F,  and  parallel  thereto  draw  E  A  meeting 
O  D  in  A  ;  make  H  C  and  E  H  each  equal  D  A,  and  upon 
the  centres  A,  B  and  C,  through  the  points  D,  H,  and  E,  let 
the  circles  A  r  D,  C  7n  H  and  B  n  E  be  described. 


PROBLEM  LVII. 

To  divide  a  given  circle  into  any  number  of  equal  parts 
by  means  of  concentric  circles. 

Let  it  be  required  to  divide  the  circle  A  K  L  into  three 
equal  parts. 

Divide  the  radius  A  B  into  three 
equal  parts  ;  and  from  the  points  of 
section  a,  d,  draw  the  perpendiculars 
a  b,  d  c,  meeting  the  circumference  of  a 
a  semicircle  described  on  A  B,  in  6 
and  c ;  and  join  B  c,  B  b.  Then,  if 
circles  be  described  from  B,  as  a  cen- 
tre, with  the  radii  B  c,B  b,  the  circle 
A  K  L  will  be  divided  into  three  equal  parts,  as  required. 


PROBLEM  LVIIL 

To  describe  the  circumference  of  a  circle  through  two  given 
points.  A,  B,  which  shall  touch  a  right  line,  C  D,  whose 
position  is  given. 

1.  Draw  the  right  line  A  B, 
joining  the  two  given  points, 
which  bisect  in  E,  by  the  perpen- 
dicular E  F,  meeting  the  given 
line  C  D  in  F. 

2.  Join   B  F,  and  from  any 
point,  H,  in  F  E,  let  fall  the  per- 
pendicular H  G ;  and  having  made  H  I  equal  to  H  G,  draw 
B  K,  parallel  to  I  H,  meeting  F  E  in  K. 

3.  Then,  if  a  circle  be  described  from  the  point  K,  as  a 
centre,  with  the  radius  K  B,  it  will  pass  through  the  pomla 
A  and  B.  and  touch  the  line  C  D.  as  was  required. 


56 


PRACTICAL    GEOMETRY. 


PROBLEM  LEC. 


Upon  a  given  line,  A  B,  to  describe  an  oval,  or  a  figure 
resembling  an  ellipse. 

1.  Divide  A  B  into  three  equal  parts, 
A  C,  C  D,  D  B  ;  and  from  the  points 
C,  D,  with  the  radii  C  A,  D  B,  describe 
the  circles  A  G  D  E,  and  C  H  B  F. 

2.  Through  the  intersections  m,  n, 
and  centres  C,  D,  draw  the  lines  tn  H, 
n  E ;  and  from  the  points  n,  m,  with  the  radii  n  E,  m  H,  de- 
scribe the  arcs  E  F,  H  G,  and  A  G  H  B  F  E  will  be  the 
oval  required. 

Another  method. 

c 

1.  On  A  B,  as  a  common  base, 

describe  the  two  equal  isosceles 
triangles  A  C  B,  A  D  B,  produc- 
ing their  sides  C  A,  C  B,  D  A,  and 
DB. 

2.  From  D  and  C,  as  centres, 
with  any  convenient  radius,  de- 
scribe the  arcs  E  G,  F  H ;  and 
from  A  and  B,  with  the  radius  A  E, 
B  G,  describe  the  arcs  G  H,  E  F,  and  the  figure  E  F  H  G 
will  be  the  oval  required. 

PROBLEM  LX. 

To  divide  a  given  straight  line  A  B  in  extreme  and  mean 
ratio ;  that  is  so  that  the  whole  line  shall  be  to  the  greater 
part  as  the  greater  part  is  to  the  less. 

1.  From  A,  draw  A  C  at  right 
angles  to  A  B,  and  make  A  C 
=  A  B ;  produce  B  A  to  D, 
making  A  D  =  half  A  B ;  and 
about  D,  at  the  distance  D  C, 
describe  the  arc  C  E.  Then 
A  B  is  divided  in  extreme  and 
mean  ratio  at  the  point  E ;  and 
AB    AE::AE:EB. 


MENSURATION  OF  SUPERFICIES. 


§  4.  The  area  of  any  figure  is  the  measure  of  its  surface, 
or  of  the  space  contained  within  the  bounds  of  that  surface, 
without  any  regard  to  the  thick- 
ness. A  square  whose  side  is  1 
inch,  1  foot,  or  1  yard,  &c.,  is 
called  the  measuring  unit ;  and  the 
area,  or  content  of  any  figure,  is 
estimated  by  the  number  of  squares 
of  this  kind  that  are  contained  in  it, 
as  in  the  rectangle,  A  B  C  D. 


•1 1 •;  —  r — ( — " 

*         ■          I         I          I 


THE  SGIUARE. 

PROBLEM  I. 

§  5-  To  find  the  area  of  a  square. 

Rule. — Multiply  the  side  by  itself,  and  the  product  will 
be  the  area. 


EXAMPLES. 

1.  What  is   the  area  of  the  square  d 
A  B  C  D,  whose  side  is  21  inches  ? 

Here,  21  x21=441  inches :  the  area 
required. 

2.  What  is  the  area  of  a  square  whose 
side  is  4  ft.  2  in.  ?      ^ns.  17  ft.  4  in.  4".   '^ 


67 


58  MENSURATION    OF    SUPERFICIES. 

m 

3.  What  is  the  area  of  a  square  field  whose  side  is  50 
perches  ?  ^ns.  15  a.  2  r.  20  p. 

4.  What  in  the  area  of  a  square  meadow  whose  side  is 
35.25  chains?  J3ns.  124  a.  1  r.  1  p. 

5.  How  many  yards  are  contained  in  a  square  whose  side 
is  3G  feet  ?  Jins.  144  yds. 

(}.  How  many  men  can  stand  on  6  acres  of  land,  each 
occupying  a  space  of  3  feet  square  ?        ^7is.  29040  men. 

PROBLEM  II. 

The  area  of  a  square  heing  given,  to  find  the  length  of  the 
side. 

Rule. — Extract  the  square  root  of  the  area. 

EXAMPLES. 

1.  The  area  of  a  square  is  2025  feet ;  what  is  the  side  ? 
Here  ^  2025  =  45  ft.  the  side  required. 

2.  What  is  the  side  of  a  square  floor  containing  729  square 
feet  ?  ^ns.  27  ft. 

3.  What  is  the  side  of  a  square  whose  area  is  8  acres, 
0  roods,  16  perches  ?  ^ns.  36  p. 

4.  What  is  the  side  of  a  square  field  whose  area  is  7  acres  ? 

^ns.  8.3666  chains. 

5.  Required  the  side  of  a  square  floor  containing  1734 
square  feet  ?  w?ns.  41.6413  ft. 

6.  Required  the  side  of  a  square  meadow  whose  area  is 
6  acres,  2  roods,  14  perches  ?  ^ns.  32.4653  p. 

PROBLEM  m. 

The  diagonal  of  a  square  Hing  given,  to  find  the  area. 

Rule. — Divide  the  square  of  the  diagonal  by  2,  and  the 
juotient  will  be  the  area. 


MENSURATION    OF    SUPERFICIES.  69 

EXAMPLES. 

1.  The  diagonal  of  thesquare,  A  BCD,  D 
IS  8  chains  ;  what  is  the  area  ? 

Here  (8x8) -7-2=64 -.-2=32  square 
chains,  then  32-=- 10=3  a.  Or.32  p. 

2.  The    diagonal  of  a  square  is   12 
yards  ;  what  is  the  area  ?    Ans.  72  yds.    eX 

3.  The  diagonal  of  a  square  is  34  perches ;  what  is  the 
area  ?  Ans.  3  a.  2  r.  18  p. 

4.  The  diagonal  of  a  square  is  16  chains ;  what  is  the 
area  ?  Ans.  12  a.  3  r.  8  p. 

5.  The  diagonal  of  a  square  is  324  feet ;  what  is  the  area  ? 

Ans.  52488  ft. 

C.  How  many  acres  are  contained  in  a  square  field  whose 
diagonal  is  29  chains  ?  Ans.  42  a.  0  r.  8  p. 


PROBLEM  IV. 

The  area  of  a  square  being  given,  to  find  the  diagonal. 
Rule. — Extract  the  square  root  of  double  the  area. 

EXAMPLES. 

1.  The  area  of  a  square  piece  of  land  is  64  acres,  3  roods, 
8  perches  ;  what  is  the  diagonal  ? 

Here  64  a.  3  r.  8  p.  =  10368  perches;  then  ^  (10368 
X2)=  v^  20736=  144  perches,  the  diagonal. 

2.  The  area  of  a  square  is  578  feet ;  what  is  the  diago- 
nal ?  Ans.  34  ft. 

3.  The  area  of  a  square  is  128  yards  ;  what  is  the  diago- 
nal ?  Ans.  16  yds. 

4.  The  area  of  a  square  field  ia  28.8  acres  ;  what  is  the 
diagonal  in  chains  ?  Ans.  24  chs. 


60  •        MENSURATION   OF   SUPERFICIES. 

5.  The  area  of  a  square  meadow  is  16.2  acres  ;  what  is 
the  diagonal  in  chains  ?  Jins.  18  chains. 

6.  The  area  of  a  square  is  4  acres  and  8  perches ;  what  is 
the  diagonal  ?  Ans.  36  perches. 


PROBLEM  V. 

The  diagonal  of  a  square  being  given,  to  find  the  side. 

Rule. — ^Extract  the  square  root  of  half  the  square  of  the 
diagonal. 

EXAMPLES. 

1.  The  diagonal  of  a  square  is  24  yards ;  what  is  the  side  ? 
Herev/[(24  x24) -r-2] = -v/(576-T-2) = v^288= 16.9705yds. 

2.  The  diagonal  of  a  square  is  18  feet;  what  is  the  side? 

Ans.  12.7279  feet. 

3.  The  diagonal  of  a  square  is  36  chains ;  what  is  the 
side  ?  Am.  25.4558  chains. 

4.  What  is  the  side  of  a  square  whose  diagonal  is  48 
perches?  Ans.  33.9411  perches. 

5.  What  is  the  side  of  a  square  whose  diagonal  is  12  feet  ? 

Jins.  8.4852  feet. 

6.  What  is  the  side  of  a  square  piece  of  land  whose 
diagonal  is  58  perches  ?  Jins.  41.0121  perches. 


PROBLEM  VL 

To  cut  off  a  given  area  from  a  square,  parallel  to  either 
side. 

Rule. — Divide  the  given  area  by  the  length  of  the  side, 
the  quotient  will  be  the  length  of  the  other  side  to  be  cut  off. 


MENSURATION   OF   SUPERFICIES. 


61 


EXAMPLES. 

1.  What  length  must  be  cut  from  the 
square  A  BCD,  whose  sides  are  25 
chains,  to  have  an  area  B  C  E  F,  of  40 
acres,  at  the  end  ? 

Here  400  ch.-j-  25  =  16  chains,  length 
required. 

2.  The  sides  of  a  square  are  17  feet ; 
what  must  be  the  length  of  another  side 
to  give  an  area  of  153  square  feet  ? 


Ans.  9  feet. 


3.  The  side  of  a  square  being  40  rods,  what  must  be  the 
length  of  the  other  side,  to  give  45  acres  ? 

Arts.  18  rods. 

4.  The  sides  of  a  square  are  15  chains  ;  what  must  be  the 
length  of  the  other  side  to  have  an  area  of  Kh.  acres  ? 

^ns.  5  chains. 

5.  What  length  must  be  cut  off  from  a  square  field  whose 
sides  are  125  perches  to  have  an  area  of  50  acres  ? 

^ns.  64  perches. 

6.  The  area  of  a  square  piece  of  land  is  29  acres,  3  r.  1  p.; 
what  length  must  be  cut  off  from  the  same  to  give  an  area 
of  10  acres  and  .8  perches  ?  ^ns.  23.2  perches. 


THE  RECTANGLE. 
PROBLEM  I. 


§  G.  The  length  and  breadth  of  a  rectangle  being  given 
w  find  the  area. 

Rule. — Multiply  the  length  by  the  breadth,  and  the 
product  will  be  the  area. 

6 


62  MENSURATION    OF    SUPERFICIES. 


EXAMPLES. 

1.  What  is  the  area  of  the  rect- 
angle A  B  C  D,  whose  length  A  B    D' 
is    14.5  feet,  and  breadth  BC  11.6 
feet? 

Here,   14.5x11.6  =  168.2  feet, 
area  required.  ^. 

2.  What  is  the  area  of  a  rectangle 
whose   length  is  14  feet  6  inches,   and  breadth  4  foet  9 
inches  ?  £ns.  68  ft.  10|  in. 

3.  How  many  acres  are  contained  in  a  rectangular  piece 
of  land  whose  sides  are  46  and  58  chains  ? 

^ns.  266  acres,  3  r.  8  p. 

4.  How  many  square  feet  are  contained  in  28  boards,  each 
18  feet  long  and  16  inches  wide  ?  *^ns.  672  feet. 

5.  How  many  squares  of  100  feet  each  are  contained  in 
a  floor  48  feet  long  and  21  feet  wide  ? 

^ns.  10.08  squares. 

6.  What  is  the  area  of  a  rectangular  piece  of  land  whose 
length  is  204.7  and  breadth  117.8  yards  ? 

^ns.  24113.66  yards. 


PROBLEM  n. 

The  area  and  either  side  of  a^ rectangle  being  given,  to  find 
the  other  side. 

Rule. — Divide   the   area   by  the   given  side,   and    the 
quotient  will  be  the  other  side. 

EXAMPLES. 

1.  The  area  of  a  rectangle  is  456  feet,  and  the  length 
30  feet ;  what  is  the  breadth  ? 

Here  456  -i-  30  =  15.2  feet,  the  breadth. 

2.  The  area  of  a  rectangle  is  846  chains,  and  its  length 
42  chains  ;  what  is  the  breadth  ?  ^ns.  204  chains, 


mensuhation  of  superficies.  63 

3.  The  area  of  a  rectangular  piece  of  land  is  6  a.  2  r.  16  p., 
and  breadth  29|  p.;  what  is  the  length  ? 

£ns.  36  perches. 

4.  The  area  of  a  rectangle  is  392  yards,  and  the  shortest 
side  12  yards ;  what  is  the  longest  side  ? 

Ans.  32f  yards. 

5.  The  area  of  a  rectangular  meadow  is  73  acres,  3  roods 
20  perches,  and  the  length  120  perches  ;  what  is  the  breadth  ? 

Ans.  98^  p. 

6.  The  area  of  a  rectangle  is  1728  feet,  and  the  breadth 
36  feet ;  what  is  the  length  I  Ans.  48  ft. 

PROBLEM  III. 

The  area  and  the  proportion  of  the  two  sides  of  a  rectangle 
being  given,  to  find  the  sides. 

Rule. — Multiply  the  area  by  the  greater  number  of  the 
proportion,  and  divide  the  product  by  the  less  ;  the  square 
root  of  the  quotient  will  be  the  length :  then  multiply  the 
length  by  the  less  number  of  the  proportion,  and  divide  the 
product  by  the  greater ;  the  quotient  will  give  the  breadth. 

EXAMPLES. 

1.  The  area  of  a  rectangular  piece  of  land  is  432  acres, 
and  the  length  is  to  the  breadth  as  5  to  3  ;  what  are  the 
sides  ? 

Here  432  acres  =  69120  perches. 

Then^/  [(6yi20x5)-T-3]=v/  (;M5600-4-3)  =  ^Z 115200 
=  339.41125  perches,  the  length  ; 
And  (339.41 125x3)^5=203.64675perches,the  breadth. 

2.  The  area  of  a  rectangle  is  1472  yards,  and  the  breadth 
is  to  the  length  as  3  to  4 ;  required  the  sides. 

Jins.  33.2264,  and  44.3019  yds. 

3.  The  area  of  a  rectangle  is  24  acres,  and  the  length  is 
to  the  breadth  as  3  to  2 ;  what  are  the  sides  ? 

Ans.  18.9736,  and  12.6491  chains. 

4.  The  area  of  a  rectangle  is  27  acres,  3  roods,  20  perches, 
and  the  length  is  to  the  breadth  as  9  to  7 ;  required  the  side^, 

•  Am.  75.725,  and  58.897  p. 


64  MENSURATION    OF    SUPERFICIES. 

5.  The  area  of  a  rectangle  is  28  acres,  and  the  breadth  it 
to  the  length  as  4  to  7 ;  required  the  sides. 

^ns.  13.6491,  and  22.1359  chains. 

6.  It  is  required  to  lay  out  1  acre  in  a  rectangular  form, 
having  the  length  to  the  breadth  in  the  ratio  of  8  to  5. 

^ns.  88,  and  55  yds. 


PROBLEM  IV. 

The  sides  of  a  rectangle  being  given,  to  cut  off  a  given  area 
parallel  to  either  side. 

Rule. — Divide  the  area  by  the  side  which  is  to  retain  its 
length  or  breadth,  and  the  quotient  will  be  the  length  oi 
breadth  of  the  other  side. 

EXAMPLES, 

1.  The  sides  of  the  rectangle, 
A  B  C  D,  are  18.16,  and  12.15 
chains  ;  what  must  be  the  length 
to  leave  an  area,  BCEF,of  13 
acres  adjoining  the  breadth  ? 

Here    120  chs.  -4-  12.15  = 
9.8765  chains,  the  length. 

2.  The  sides  of  a  rectangle  are  34  and  1 6  chains ;  what 
must  be  the  breadth  to  leave  18  acres  adjoining  the  length  ? 

jins.  7i  chs. 

3.  The  sides  of  a  rectangle  are  216  and  124  feet ;  what 
must  be  the  length  to  leave  10106  square  feet  adjoining  the 
breadth?  ^n*.  8U  ft. 

4.  The  sides  of  a  rectangle  are  180  and  75  perches  ;  what 
must  be  the  breadth  so  as  to  leave  22^  acres  adjoining  the 
length  ?  ^ns.  20  p. 

5.  The  sides  of  a  rectangle  are  8  and  15  yards ;  what 
must  be  the  length  to  leave  46  square  yards  at  the  shortest 
side  ?  ^ns.  5?  yds. 

6.  The  length  of  a  rectangle  is  14.5  chains,  and  the  breadth 
6.4  chains  ;  what  must  be  the  breadth,  the  length  being  the 
same,  to  contain  5.8  acres  ?  jirm.  4  chs. 


MENSURATION    OF   SUPERnCIES.  65 


.  THE  RHOMBUS. 

PROBLEM  I. 

§  7.  To  find  the  area  of  a  rhombus. 

Rttle. — Multipl}'  the  length  by  the  perpendicular  height, 
and  the  product  will  be  the  area. 

EXAMPLES. 

1.  The  length  of  a  rhombus,  A  B, 
is  I82  feet,  and  the  perpendicular 
height,  D  E,  7|  feet ;  required  the 
area.  i 

Here,  18.5x7.75=143.375  feet.  / 


2.  Required  the  area  of  a  rhombus      a/ 
whose  length  is  14  feet  4  inches,  and 
its  height  12  feet  2  inches. 

Ans.  174  ft.  4  in.  8". 

3.  What  is  the  area  of  a  rhombus  whose  length  is  12  feet 
3  inches,  and  height  9  feet  4  inches  ?     Ans.  114  ft.  4  in. 

4.  Required  the  area  of  a  rhombus  whose  length  is  38 
perches,  and  height  17  perches.  Ans.  4  a.  0  r.  6  p. 

5.  What  is  the  area  of  a  rhombus  whose  length  is  19 
chains,  and  height  15  chains  ?  Ans.  28  a.  2  r. 

6.  Required  the  area  of  a  rhombus  whose  length  is  27 
yards,  and  height  21.5  yards.  Ans.  580.5  yds. 


G« 


^6  MENSURATION    OF    SUPRRFICIES. 

THE  RHOMBOID. 

PROBLEM  I. 

§  8.  To  find  the  area  of  a  rhomboid. 

Rule. — Multiply  the  length  by  the  perpendicular  height, 
and  the  product  will  be  the  area. 

EXAMPLES. 

1.  The  length  of  a  rhomboid,  A  B, 
is  28  perches  and  height,  D  E,  12  ; 
required  the  area. 

Here  28  X 12 =336  perches =2  a. 
0  r.  16  p. 

2.  The  length  of  a  rhomboid  is 
22  feet  9  inches,  and  height  14  feet 

3  inches  ;  how  many  square  yards  does  it  contain  ? 

^ns.  36.0208  yds. 

3.  The  length  of  a  rhomboid  is  16.2  yards,  and  height 
9.6  yards  ;  how  many  square  perches  does  it  contain  ? 

Ans.  5.1411  sq.  p. 

4.  The  length  of  a  rhomboid  is  21  chains,  and  the  height 
11.5  chains  ;  what  is  the  area  ?         Ans.  24  a.  0  r.  24  p. 

5.  How  many  acres  are  contained  in  a  rhomboid  whose 
length  is  130  perches  and  height  57  perches  ? 

Ans.  46  a.  1  r.  10  p. 

6.  How  many  square  yards  are  contained  in  a  rhomboid 
whose  length  is  271  feet  and  height  107  feet  ? 

Ans.  322 If  sq.  yds. 

PROBLEM  IL 

The  area  of  a  rhombus  or  rhomboid,  and  the  length  of  the 
side  being  given,  to  find  the  perpendicular  height ;  or  the 
area  and  the  height  being  given,  to  find  the  length  of  the 
side. 

Role. — Divide  the  area  bv  the  length  of  the  side,  and  the 
quotient  will  be  the  perpenaicular  height ;  or  divide  by  the 
height,  and  the  quotient  will  be  the  length  of  the  side. 


MENSURATION    OF    SUPERFICIES.  67 


EXAMPLES. 

1.  The  area  of  a  rhombus  is  27  perches,  and  the  length 
of  the  side  6.75  perches;  what  is  the  perpendicular  height  I 

Here  27  ■—  6.75  =  4  perches. 

2.  The  area  of   a  rhombus  is  4  acres,  and  the  height 
5  chains ;  what  is  the  length  of  the  side  ? 

^ns.  8  chains. 

3.  Required  the  length  of  a  rhombus  whose  area  is  17 
acres,  and  height  35  perches.  ^ns.  77^  perches. 

4.  The  area  of  a  rhomboid  is  4  acres,  3  roods,  18  perches, 
and  the  length  of  the  side  38  perches ;  what  is  the  height  ? 

^ns.  20^*9  perches. 

5.  The  area  of  a  rhomboid  is  1776  square  feet,  and  the 
height  24  feet ;  what  is  the  length  ?  ^ns.  74  feet. 

6.  The  area  of  a  rhomboid  is  36  acres,  and  the  length  of 
the  sides  24  chains ;  what  is  the  height  ? 

^ns.  15  chains. 


THE  TRIANGLE. 
PROBLEM  I. 


§  9.  To  find  the  area  of  a  triangle,  when  the  base  and 
perpendicular  height  are  given. 

Rule. — Multiply  the  base  by  the  perpendicular  height, 
and  half  the  product  will  be  the  area. 

EXAMPLES. 

1.  What  is  the  area  of  a  triangle 
ABC,  whose  base,  AB,  is  46  feet, 
and  height,  DC,  23  feet  ? 

Here  (46  x  23)  ^2  =  1058-}-2= 
529  feet,  the  area. 

2.  What  is  the  area  of  a  triangle 
whose  base  is  67  yards,  and  height 
is  14.5  yards  ?       Jins.  485.75  yds. 


68  MENSURATION    OF    SUPERFICIES. 

3.  What  is  the  area  of  a  triangle  whose  base  is  207.5  feet, 
and  height  is  59.5  feet  ?  Ans.  6173.125  feet. 

4.  The  base  of  a  triangle  is  72  perches,  and  the  height  122 
perches ;  how  many  acres  does  it  contain  ? 

Ans.  2  a.  3  r.  10  p. 

5.  What  is  the  area  of  a  triangle  whose  base  is  12.25 
chains,  and  the  height  8.5  chains  1      Ans.  5  a.  0  r.  351  p. 

6.  What  is  the  area  of  a  triangular  field  whose  base  is  24^ 
chains,  and  the  height  18  chains  ?       Ans.  22  a.  0  r.  8  p. 


PROBLEM  II. 
The  three  sides  of  a  triangle  being  given,  to  find  the  area. 

Rule. — From  half  the  sum  of  the  three  sides,  subtract 
each  side  severally.  Then  multiply  the  half  sum,  and  the 
three  remainders  continually  together,  and  the  square  root 
of  the  product  will  be  the  area  required. 

EXAMPLES. 

1.  What  is  the  area  of  a  triangle 
ABC,  whose  three  sides,  B C,  C  A, 
AB,  are  23.7,  29.25  and  40.1 
yards  ? 

Here  (23.7  +  29.25  +  40.1)  -^  2 
=  93.05  -5-  2  =  46.525  =  half  the 
sum  of  the  sides. 

46.525  —  23.7    =  22.825  =  first  diffejence. 
46.525  —29.25  =  17.275  =  second  difference. 
46.525  —40.1    =   6.425  =  third  difference. 
Whence      ^/  (46.525  x  22.825  x  17.275  x  6.425) 
=  ^117865.94866835  =  343.3161  yards,  the  area. 

.  2.  What  is  the  area  of  a  triangular  field  whose  sides  ate 
26,  28  and  30  chains  ?  Jlns.  33  a.  2  r.  16  p. 

3  Required  the  area  of  an  equilateral  triangle  whose  side 
is  22  perches.  Ans.  1  a.  1  r.  9.5781  p. 

4.  Required  the  area  of  an  isosceles  triangle  whose  base  is 
HO,  and  each  of  its  eaual  sides  45  feet  ? 

Ans.  636.3961  feet. 


MENSURATION    OF    SUPERFICIES.  69 

5.  What  is  the  area  of  a  triangle  whose  sides  are  22.2, 38, 
and  40.1  feet  ?  ^ns.  413.7114  feet. 

6.  What  is  the  area  of  a  triangular  field  whose  sides  are 
27.35,  31.15,  and  38  chains  ?     ^ns.  42  a.  0  r.  6.6955  p. 


PROBLEM  m. 

Any  two  sides  of  a  right-angled  triangle  being  given,  to 
find  the  third  side. 

Rule  1st. — To  the  square  of  the  perpendicular  add  the 
square  of  the  base,  and  the  square  root  of  the  sum  will  give 
the  hypothenuse. 

2d. — The  square  root  of  the  difference  of  the  squares  of  the 
hypothenuse,  and  either  side  will  give  the  other. 

3d.  Or  multiply  the  sum  of  the  hypothenuse  and  either 
side  by  their  difference,  and  the  square  root  of  the  product 
will  give  the  other. 

EXAMPLES. 

1.  The  base  of  a  right-angled  tri- 
angle, A  B,  is  27  yards,  and  the  per- 
pendicular, B  C,  36  yards;  what  is 
the  hypothenuse? 

Here  ^/  (36"  -f  27=^)  =  ^  (1296 
-I-  729)  =  v/  2025  =  45  yards,  the 
hypothenuse. 

2.  The  hypothenuse  of  a  right-angled  triangle  is  242  feet, 
and  the  perpendicular  182  feet ;  what  is  the  base  ? 

Here  v/  (242-"— 182")  =  ^  (58564—33124)  =  ^Z  25440 
=  159.4989  feet,  the  base. 

Or,  v/  [  (242  -f  182)  x  (242  —  182)  ]  =  x/  (424  x  60) 
=  v/  25440  =  159.4989  feet,  the  base  as  before. 

3.  The  base  of  a  right-angled  triangle  is  38  chains,  and 
tho  perpendicular  41  chains  ;  required  the  hypothenuse. 

^ns.  55.9016  chs 

4.  The  hypothenuse  of  a  right-angled  triangle  is  68 
perches,  and  the  perpendicular  33  perches ;  what  is  the  base  ? 

^ns.  59.4558  p. 


70  MENSURATION    OF    SUPERFICIES. 

5.  The  hypothenuse  of  a  right-angled  triangle  is  315  feet, 
and  the  base  299  feet ;  required  the  perpendicular. 

^ns.  99.1160  feet. 

6.  The  top  of  a  May  pole,  being  broken  off  by  a  blast  of 
wind,  struck  the  ground  at  9  feet  distance  from  the  foot  of 
the  pole ;  what  was  the  height  of  the  whole  May  pole,  sup- 
posing the  length  of  the  broken  piece  to  be  41  feet  ? 

£ns.  81  ft. 

7.  Two  ships  sail  from  the  same  port,  one  east  60  miles, 
and  the  other  north  80  miles ;  how  far  are  they  apart  ? 

^ns.  100  m. 

8.  A  line  78  yards  long  will  reach  from  the  top  of  a  fort, 
on  the  opposite  bank  of  a  river,  to  the  water  edge  on  this  side 
of  the  river ;  what  is  the  height  of  the  fort,  the  river  being 
76  yards  across  ?  ^ns.  17.5499  yds. 

9.  A  ladder  of  100  feet  in  length  was  placed  against  a 
building  of  100  feet  high,  in  such  a  manner  that  the  top  of 
it  reached  within  six  inches  of  the  top  of  the  building ;  what 
was  the  distance  of  the  foot  of  the  ladder  from  the  base  of  the 
edifice  ?  ^ns.  9.9874  ft. 

10.  A  ladder  30  feet  long,  placed  near  the  middle  of  a 
street,  reached  the  buildings  at  one  side  24  feet  from  the 
ground  ;  and  the  opposite  side,  without  moving  the  foot,  18 
feet :  what  was  the  breadth  of  the  street  ?  I^ns.  42  ft. 


PROBLEM  IV. 

The  sum  of  the  hypothenuse  and  perpendicular,  and  the 
lase  of  a  right-angled  triangle  being  given,  to  find  the  hypo- 
thenuse and  the  perpendicular. 

Rule. — To  the  square  of  the  sum  add  the  square  of  the 
base,  and  divide  the  number  by  twice  the  sum  of  the  hypo- 
thenuse and  perpendicular,  and  the  quotient  will  be  the 
hypothenuse. 

Subtract  the  hypothenuse  from  the  sum  of  the  hypothenuse 
and  perpendicular,  and  the  remainder  will  be  the  perpen- 
dicular. 


MENSURATION    OF    SUPERFICIES.  71 


EXAMPLES. 

1.  The  sum  of  the  hypothenuse 
and  perpendicular  is  100  feet,  and 
the  base  40  feet ;  required  the  hypo- 
thenuse and  the  perpendicular. 

Here  (100^  +  40^)  -4-  (100  x  2) 
=  (10000  +    1600)   -4-   200  = 
1 1000  -4-  200  =  58  feet,  the  hypo- 
thenuse.    And  100  —  58=  42  feet,  the  perpendicular. 

2.  The  height  of  a  tree,  standing  perpendicularly  on  a 
plane,  is  1 10  feet.  At  what  height  must  it  break  off,  so  that 
the  top  may  rest  on  the  ground  50  feet  from  the  base,  and 
the  place  broken  on  the  upright  part  ?  Ans.  43y''-j-  ft. 

3.  The  base  of  a  right-angled  triangle  is  36  chains,  and 
the  sum  of  the  hypothenuse  and  perpendicular  is  84  chains  ; 
what  is  the  hypothenuse  ?  Ans.  49f  ch. 

4.  A  tree  90  feet  high,  growing  perpendicularly  on  a 
plane,  was  broken  off  by  the  wind  ;  the  broken  part  resting 
on  the  upright,  and  the  top  on  the  ground  30  feet  from  the 
base  ;  what  was  the  length  of  the  broken  part  ? 

Ans.  50  ft. 

5.  The  sum  of  the  hypothenuse  and  perpendicular  is  240 
yards,  and  the  base  80  yards;   required  the  perpendicular. 

Jins.  106f  yds. 

6.  A  May  pole,  whose  height  was  84  feet,  standing  on  a 
horizontal  plane,  was  broken  by  the  wind,  and  the  ex- 
tremity of  the  top  part  struck  the  ground  at  the  distance  ot 
24  feet  from  the  bottom  of  the  pole  ;  required  the  length  of 
each  part. 

Ans.  45|^  feet,  the  hypolh.  and  38f  the  perp. 


PROBLEM  V. 

To  determine  the  area  of  an  equilateral  triangle. 

Rule. — Multiply  the  square  of  the  side  by  .433013,  and 
the  product  will  be  the  area. 


72 


MENSURATION    OF    SUPERFICIES. 


EXAMPLES. 

1.  What  is  the  area  of  an  equilate- 
ral triangle,  ABC,  whose  side  is  20 
feet? 

Here  (20) «  x  .433013  =  400 
X  .433013  =  173.2052  feet,  the 
area.  a 

2.  What  is  the  area  of  an  equilateral  triangle  whose  side  is 
40  feet  ?  Ans.  692.8208  feet. 

3.  What  is  the  area  of  an  equilateral  triangle  whose  side  is 
80  perches  ?  Ans.  17.3205  acres. 

4.  What  is  the  area  of  an  equilateral  triangle  whose  side 
is  24.4  yards  ?  Ans.  257.7986  yards. 

5.  How  many  acres  are  contained  in  an  equilateral  triangle 
whose  side  is  16  chains  ?  Ans.  11  a.  0  r.  13.6212  p. 

6.  How  many  acres  are  contained  in  an  equilateral  triangle 
whose  side  is  32  perches  ?  Ans.  2  a.  3  r.  3.4053  p. 


PROBLEM  VI. 

The  area  and  the  base  of  any  triangle  being  given,  to  find 
the  perpendicular  height. 

Or,  the  area  and  height  being  given,  to  find  the  base. 

Rule. — Divide  double  the  area  by  the  base,  and  the  quo- 
tient will  be  the  perpendicular  height ;  or  divide  double  the 
area  by  the  height,  and  the  quotient  will  be  the  base. 


EXAMPLES. 

1.  The  area  of  a  triangle,  ABC, 
is  4  a.  3  r.  32  p.,  and  the  base,  A  B, 
24  perches;  what  is  the  perpen- 
dicular height,  DC? 

Here  4  a.  3  r.  32  p.  =  792  p. 
and  (792  x  2)  -h  24  =  1584  -^  24 
=  66  perches,  the  height. 


MENSURATION    OF   SUPERFICIES.  73 

2.  The  area  of  a  triangle  is  806.3125  yards,  and  the  per- 
pendicular 33.25  yards ;  what  is  the  base  ? 

(806.3125  X  2)  -J-  33.25  =  1612.625  -r-  33.25  =  48.5 

yards,  the  base. 

3.  The  area  of  a  triangle  is  1025  feet,  and  the  base  20 
feet ;  what  is  the  perpendicular  ?  Ans.  10^  feet. 

4.  The  area  of  a  triangle  is  5  a.  0  r.  33  p.  and  the  per- 
pendicular 28.5  p. ;  what  is  the  base  ?  Ans.  58.4561  perches. 

5.  If  the  area  of  a  triangle  be  862A|  yards,  and  its  base 
17}  yards  ;  what  is  its  perpendicular  height  ? 

Ans.  lOOf  yards. 

6.  If  the  area  of  a  triangular  field  be  4.39775  acres,  and 
its  perpendicular  7.18  chains ;  what  is  its  base  ? 

Ans.  12.25  chains. 


PROBLEM  VII 

The  proportion  of  the  three  sides  of  a  triangle  being  given, 
to  find  the  sides  of  a  triangle  corresponding  with  a  given  area. 

Rule. — Find  the  area  of  the  triangle  according  to  the 
given  proportion,  by  Problem  II.  page  68  ;  then  as  that 
area  is  to  the  area  given,  so  is  the  square  of  either  of  its 
sides  to  the  square  of  the  similar  side ;  the  square  root  of 
which  will  be  the  required  side.  The  other  sides  of  the 
triangle  will  be  proportional  to  the  corresponding  given  sides. 

EXAMPLES. 

1.  A  person  wishes  to  enclose 
6  acres,  1  rood,  12  perches,  in  a 
triangle,  similar  to  a  small  triangle 
whose  sides  are  6,  8,  and  9  perches 
respectively  ;  required  the  sides  of 
the  triangle. 

Here  23.5252  perches  =  the  area 
of  the  triangle,  whose  sides  are  6,  8,  and  9. 

6  a.  1  r.  12  p.  =  1012  perches. 

As  23.,5252  :  1012  ::  64  :  2753.1328,  and  ^  2753.1328 
e=  h'iAll  perches,  one  of  the  sides. 

Now  as  8  :  9  : :  52.47  :  59.029  p.  ?  ,,       ,.       ,         . ,_ 
"     "8:6::  52.47  :  39.353  p.  $  ^^^  °*^^^  ^^°  '^'^^• 
7 


74  MENSURATION    OF   SUPERFluiES. 

2.  The  area  of  a  triangle  is  24  acres ;  what  must  be  the 
length  of  the  sides,  in  the  proportion  3,  4  and  5  chains  ? 

^ns.  18.973,  25.298,  and  31.622  chains. 

3.  The  area  of  a  triangle  is  10  acres ;  what  must  be  the 
length  of  the  sides,  in  the  proportion  of  5,  11  and  13  chains  ? 

.^ns.  9.G4,  21.208  and  25.004  chains. 

4.  What  are  the  sides  of  a  triangle  containing  one  acre, 
in  the  proportion  of  3,  4  and  0  chains? 

^ns.  4.1082,  5.4776  and  8.2164  chains. 

5.  What  are  the  sides  of  a  triangle  containing  33.6  acres, 
in  the  proportion  of  13,  14  and  15  chains  ? 

./ins.  26,  28  and  30  chains. 

6.  What  are  the  sides  of  a  triangle  containing  24  acres,  in 
the  proportion  of  5,  12  and  13  chains  ? 

Jns.  14.1421,  33.9411  and  36.7695  chains. 


PROBLEM  VIII. 

The  base  and  perpendicular  of  any  plane  triangle  being 
given,  to  find  the  side  of  its  inscribed  square. 

Rule. — Divide  the  product  of  the  base  and  perpendicular 
by  their  sum,  and  the  quotient  will  be  the  side  of  the  in- 
scribed square 

EXAMPLES. 

1 .  The  base  A  B  of  a  triangle  is 
12  feet,  and  the  perpendicular,  C  D, 
18  feet;  what  is  the  side,  EF,  of 
the  inscribed  square  ? 

Here  (12  x  18)  -r-  (12  +  18) 
=  216  -V-  30  =  7.2  feet,  =  EF, 
the  side  of  the  inscribed  square 

2.  The  base  of  an  isosceles  triangle  is  24  yards,  and  the 
perpendicular  16  yards ;  what  is  the  side  of  the  inscribed 
square  ?  ^ns.  9.6  yards. 

3.  The  hypothenuse  of  a  right-angled  triangle  is  16 
perches,  and  the  base  14  perches ;  what  is  the  side  of  the 
inscribed  square  ?  Ans.  4.98C8  p. 


MENSURATION    OF   SUPERFICIES.  75 

4.  The  base  of  a  scalene  triangle  is  30  chains,  and  the 
pofpcndicular  15  chains;  what  is  the  side  of  the  inscribed 
Bqaare  ?  Ans.  8f  chs. 

6.  The  area  of  a  scalene  triangle  is  32  acres,  and  the  base 
25  chains  ;  required  the  side  of  the  inscribed  square. 

Ans.  12.6482  chs. 

6.  The  area  of  an  isosceles  triangle  is  324  feet,  and  the 
perivendicular  18  feet ;  required  the  side  of  the  inscribed 
square.  Jins.  12  ft. 

PROBLEM  IX. 

Tile  three  sides  of  any  triangle  being  given,  to  find  the 
lengih  of  a  perpendicular  which  will  divide  it  into  two  right- 
angled  triangles. 

RutE. — Upon  the  base,  let  fall  a  perpendicular  from  the 
opposite  angle  ;  this  perpendicular  will  divide  the  base  into 
two  parts  called  segments,  and  the  whole  triangle  into  two 
right-angled  triangles. 

Then,  as  the  base  or  sum  of  the  segments  is  to  the  sum  of 
the  other  two  sides,  so  is  the  difference  of  these  sides  to  the 
difference  of  the  segments  of  the  base. 

To  half  the  base  add  half  the  difference  of  the  segments, 
and  the  sum  will  be  the  greater  segment ;  also  from  half  the 
base  subtract  half  the  difference  of  the  segments,  and  the  re- 
mainder will  be  the  less  segment. 

When  the  perpendicular  falls  without  the  triangle,  the 
base  is  to  ihe  sum  of  the  sides  as  the  difference  of  the  sides 
is  to  the  sum  of  the  segments  of  the  base. 

Then  in  each  of  the  two  right-angled  triangles,  ihere  will 
be  known  fiie  hypothenuse  and  base  ;  consequently,  the  per- 
pendicular luay  be  found  by  Problem  III.,  page  69. 

EXAMPLES. 

1.  In  tho  triangle,  ABC,  are 
liven,  A  B  2»s  A  C  18,  and  B  C  14 
yards ;  requir«>d  the  perpendicular, 
C  D. 

Here,  As  20:  (18 -f  14)::  (18  — 
14)  :  6.4,  the  diflerence  of  the  seg- 
ments of  the  bawe  :  then,  6.4 -r- 2  = 
3.2  yards,  half  tieir  difference,  and 


76  MENSURATION    OF   SUPERFICIES. 

20  ^  2  =  10  yards,  half  the  base  ;  now  10  +  3.2  =  lli.:i 
yards,  the  segment  A  D,  and  10  —  3.2  =  G.8  yards,  the  seg- 
ment B  D.  Therefore,  ^/  (B  C^  —  B  D^)  =  -v/  ( 14«  —  6.8'^) 
=  >/  (196.  —  46.24)  =  -v/  149.76  =  12.2376  yards,  the 
perpendicular,  D  C 

2.  The  base  of  a  triangle  is  426  feet,  and  the  other  two 
sides  365  and  230  feet ;  required  the  length  of  the  perpen- 
dicular, ^ns.  196.9904  ft. 

3.  The  base  of  a  triangle  is  80  chains,  and  the  other  two 
sides  60  and  40  chains  ;  what  is  the  length  of  the  perpen- 
dicular which  divides  it  into  two  right-angled  triangles  ? 

Jns.  29.0473  chs. 

4.  The  base  of  a  triangle  is  128  perches,  and  the  other 
two  sides  94  and  68  perches ;  required  the  length  of  the 
perpendicular.  .^ns.  48.6137  p. 

5.  The  base  of  a  triangle  is  324  yards,  and  the  other  two 
sides  264  and  162  yards  ;  required  the  length  of  the  perpen. 
dicular.  ^ns.  131.2613  yds. 

6.  The  base  of  a  triangle  is  30  perches,  and  the  other  two 
sides  24  and  18  perches  ;  required  the  length  of  the  perpen- 
dicular, ^ns.  14.4  p. 

PROBLEM  X. 

The  area  and  base  of  a  triangle  being  given,  to  cut  off  a 
given  part  of  the  area  by  a  line  running  from  the  angle  oppo- 
site the  base. 

Rule. — As  the  given  area  of  the  triangle  is  to  the  area  of 
the  part  to  be  cut  off,  so  is  the  given  base  to  the  base  corre- 
sponding to  that  area. 

EXAMPLES. 

1.  Given  the  area  of  a  triangle, 
A  B  C,  12  acres,  and  the  length  of 
the  base,  A  B,  24  chains ;  it  is  re- 
quired to  cut  off  5  acres  towards  the 
angle  A,  by  a  line  running  from  the 
angle  C  to  the  base. 

Here  12  acres  =  120  chains,  and 
6  acres  =  50  chains. 


MENSURATION    OF    SUPERFICIES.  77 

Then,  as  A  B  C  (120) :  A  D  C  (50)  : :  A  B  (24) :  A  D 
(10  chains.) 

2.  In  the  triangle  ABC,  there  are  given  the  area  54 
acres,  2  roods,  30  perches,  and  the  base,  A  B,  70  perches, 
to  cut  off  20  acres  towards  the  angle  B,  by  a  hne,  C  D,  run- 
ning from  the  angle  C  to  the  base  ;  required  the  part  BD  of 
the  base.  Ans.  25.0  p. 

3.  In  the  triangle  ABC,  there  are  given  the  area  7a 
acres,  and  the  base,  A  B,  8  chains,  to  cut  off  1  g  acres  towards 
the  angle  A,  by  a  line,  C  D,  running  from  the  angle  C  to  the 
base  ;  required  the  part  A  D  of  the  base.  Ans.  2  chs. 


PROBLEM  XI. 

The  area  and  base  of  a  triangle  being  given,  to  cut  off  a 
triangle  containing  a  given  area,  by  a  line  running  parallel 
to  one  of  its  sides. 

Rule. — As  the  given  area  of  the  triangle  is  to  the  area 
of  the  triangle  to  be  cut  off,  so  is  the  square  of  the  given  base 
to  the  square  of  the  required  base.  The  square  root  of  the 
result  will  be  the  base  of  the  required  triangle. 

EXAMPLES. 

1.  Given  the  area  of  the  triangle 
ABC,  250  square  chains,  and  the 
base,  A  B,  20  chains  ;  it  is  required 
to  cut  off  60  square  chains  towards 
the  angle  A,  by  a  hne,  D  E,  running 
parallel  to  B  C. 

As  A  B  C  (250) :  A  D  E  (60) : : 
A  B^'  (400)  :  A  D^^  (96).  And  A  D 
=  v/  A  D^  (06)  =  9.7979  chains. 

2.  Given  the  area  of  a  triangle,  A  B  C,  20  acres,  and  the 
base,  A  B,  50  chains,  to  find  D  B,  a  part  of  the  base,  so  that 
a  line,  D  E,  running  from  the  point  D,  parallel  to  the  side. 
A  C,  may  cut  off  a  triangle,  B  D  E,  containing  9  acres. 

Jins.  B  D  =  33.541  chs. 

3.  Given  the  area  of  a  triangle,  A  B  C,  5  acres,  and  the 


78  MENSURATION    OF    SUPLRFICIES. 

base,  A  B,  I2h  perches,  to  find  A  D,  a  part  of  the  base,  so 
that  a  line,  D  E,  running  from  the  point  D,  parallel  to  the 
side  B  C,  may  cut  off  a  triangle,  A  D  E,  containing  2|  acres. 

^ns.  A  D  =  8.3852  p. 


PROBLEM  XII. 

The  area  and  two  sides  of  a  triangle  being  given,  to  cut 
off  a  triangle  containing  a  given  area,  by  a  line  running  from 
a  given  point  in  one  of  the  given  sides,  and  falling  on  the 
other. 

Rule. — As  the  given  area  of  a  triangle  is  to  the  area  of 
the  part  to  be  cut  off,  so  is  the  rectangle  of  the  given  sides  to 
a  fourth  term. 

Divide  this  fourth  term  by  the  distance  of  the  given  point 
from  the  angular  point  of  the  two  given  sides  ;  the  quotient 
will  be  the  distance  of  the  required  point  from  the  same 
angle. 

EXAMPLES. 

1.  Given  the  area  of  a  triangle, 
A  B  C,  2^  acres  ;  the  side  A  B,  25 
perches ;  the  side  A  C,  20  perches ; 
and  the  distance  of  a  point  D,  from 
the  angle  A,  18  perches  ;  it  is  re- 
quired to  find  a  point,  E,  to  which, 
if  a  line  be  drawn  from  the  point  D, 
it  shall  cut  off  a  triangle,  A  D  E, 
containing  1  acre,  2  roods,  10  perches. 

Here,  as  the  area  of  the  triangle  ABC,  400  sq.  p. :  the 
area  of  the  triangle  A  D  E,  250  sq.  p. : :  A  B  x  A  C  (25  x 
20) :  A  D  X  A  E  (312.5).  Then  A  D  x  A  E  (312.5)  -^ 
A  D  (18)  =  A  E  =  312.5  -4-  18  =  17.36  per. 

2.  Given  the  area  of  a  triangle,  ABC,  24.7875  acres  ; 
the  side  A  B,  40  chains ;  the  side  A  C,  .32.5  chains ;  and 
the  distance  of  a  point  E,  in  the  side  A  B,  from  the  angle  A. 
17  chains ;  it  is  required  to  find  the  distance  A  D,  in  the 
line  A  C,  so  that  a  line  drawn  from  E  to  D  may  cut  off  a 
triangle  A  E  D,  containing  6  acres.  .^ns.  18.51  chs. 


MKNSORATION    OF    SUPKRFICIES.  7? 

THE   TRAPEZIUM. 

PROBLEM  I. 

§  10.  To  find  the  area  of  a  trapezium. 

Rule. — Multiply  the  sum  of  the  two  perpendiculars  by 
the  diagonal  upon  which  they  fall,  from  the  opposite  angl(3s 
and  half  the  product  will  be  the  area. 

EXAMPLES. 

1.  Required  the  area  of  the  trape- 
zium, A  C  B  D,  whose  diagonal,  A  B, 
is  84  yards,  the  perpendicular  C  E, 
28  yards,  and  the  perpendicular  D  F, 
21  yards. 

Here  [(28  +  21)  x  84]  -4-  2  = 
(49  X  84)  -4-  2  =  41 16  -T-  2  =  2058  yards,  the  area  of  the 
trapezium  A  C  B  D. 

2.  Required  the  area  of  a  trapezium  whose  diagonal  is  3;i 
perches,  and  the  perpendiculars  11  perches  and  18  perches. 

Ans.  2  a.  1  r.  36  p. 

3.  How  many  acres  are  there  in  the  trapezium  whose 
diagonal  is  80.5  chains,  and  the  perpendiculars  22.4  and 
28.3  chains  ?  .ins.  204  a.  0  r.  10.8  p. 

4.  What  is  the  area  of  a  trapezium  whose  diagonal  is  108 
feet  6  inches,  and  the  perpendiculars  56  feet  3  inches  and 
60  feet  9  inches  ?  Ajns.  6347.25  ft. 

5.  How  many  square  yards  of  paving  are  there  in  a  trape- 
zium  whose  diagonal  is  65  feet,  and  the  two  perpendiculars 
let  fall  on  it,  from  its  opposite  angles,  28  and  335  feet  re- 
spectively ?  Ans.  222.083  yds. 

6.  How  many  acres  are  there  in  the  trapezium  whose 
diagonal  is  4.75  chains,  and  the  two  perpendiculars  falling 
on  it,  from  its  opposite  angles,  2.25  and  3.6  chains  respect- 
ively ?  Ans.  1  a.  1  r.  22.3  p. 


80  MENSURATION   OF   SUPEKFICIES. 

THE  TRAPEZOID. 
PROBLEM  I. 

§  1 1,  To  find  the  area  of  a  trapezoid. 

KfiLE. — Multiply  the  sum  of  the  two  parallel  sides  by  the 
perpendicular  distance  between  them,  and  half  the  product 
will  be  the  area. 

EXAMPLES. 

1.  Required  the  area  of  the  trape-  d c  _ 

zoid,  A  B  C  D,  whose  parallel  sides,  /\\        V 

D  C  and  A  B,  are  14  feet  (3  inches  /  |    N.       v, 

and  24  feet  9  inches,  and  the  perpen-  /     !       Ny  > 

dicular  distance,  D  E,  8  feet  3  inches.  /       I           \. 

Here  [(A  B  +  D  C)  x  D  E]  -f-  2     /i_ ' 

=  [(24.75  +  14.5)  X  8.25]  -^  2  =      ^  ^  " 

(39.25  X  8.25)  -i-  2  =  323.8125  -r-  2  =  IGl. 90625  feet,  the 
area. 

2.  How  many  square  feet  are  there  in  a  plank  1  foot  6 
inches  broad  at  one  end,  and  1  foot  3  inches  at  the  other,  the 
length  being  20  feet  ?  Mns.  27|  ft. 

3.  Required  the  area  of  a  trapezoid  whose  parallel  sides  are 
24.46  chains,  and  38.4  chains,  and  the  perpendicular  dis- 
tance 16.2  chains.  ^ns.  50  a.  3  r.  2f).6  p. 

4.  Required  the  area  of  a  trapezoid  whose  two  parallel 
sides  are  25  feet  6  inches,  and  18  feet  9  inches,  and  the  per- 
pendicular distance  between  them  10  feet  5  inches. 

^ns.  2301 1  ft. 

5.  The  two  parallel  sides  of  a  trapezoid  arb  12.41  and 
S.22  chains,  respectively,  and  the  perpendicular  distance 
between  them  5.15chains  ;  required  the  area. 

^"ins.  5  a.  1  r.  9.956  p. 

6.  Required  the  area  of  a  trapezoid  whose  two  parallel 
sides  are  750  and  1225  links,  and  the  perpendicular  distance 
between  them  1540  links.  Jlns.  15  a.  0  r.  33.2  p 


MENSURATION    OF   SUPERFICIES  81 

POLYGONS. 

PROBLEM  I. 
§  12,  To  find  the  area  of  a  regular  polygon. 

Rule. — Multiply  the  perimeter,  or  sum  of  all  the  sides  of 
the  figure,  by  the  perpendicular  falling  from  its  centre  upon 
one  of  the  sides,  and  half  the  product  will  be  the  area. 

EXAMPLES. 

1.  Required  the  area  of  the  regular 
pentagon,  ABODE,  one  of  whose 
equal  sides,  AB,  or  BC,  &«.,  is  25 
yards,  and  the  perpendicular  O  P  from 
its  centre,  17.2  yards. 

Here  (25  x  5  x  17.2)  -f-  3  r«  2150 
-5-  2  ^  1075  yards,  the  area. 

2.  The  side  of  a  regular  heiaron  is  15  feet,  and  the  per- 
pendicular 13  feet ;  what  is  the  area  ?         ^ns.  585  feet. 

3.  Required  the  area  of  a  regtiliv  hexagon  whose  side  is 
14.6  yards,  and  perpendicular  12.01  yards  ? 

Jlns.  553.632  yds. 

4.  How  many  acres  are  containert  i**!  a  regular  heptagon 
whose  sides  are  each  19.38  chains,  i>.nd  «)erpendicular  from 
the  centre  20  chains  ?  ,%is.  135.66  acres. 

5.  The  side  of  a  regular  pentagon  is  TO  yards,  and  the 
perpendicular  is  6.882  yards  ;  what  is  the  aroa  ? 

Jins.  172.%  yards. 

6.  What  IS  the  area  of  a  regular  octagon  whose  aides  are 
each  19.882  feet,  and  the  perpendicular  from  the  crc^ro  24 
feet  ?  ^ns.  1908.672  fe-* 


82 


MENSURATION    OF   SUPERFICIES. 


PROBLEM  II. 

To  find  the  area  of  a  regular  polygon  when  one  of  its 
equal  sides  only  is  given. 

Rule. — Multiply  the  square  of  the  side  of  the  polygon, 
by  the  number  standing  opposite  to  its  name  in  the  following 
table,  and  the  product  will  be  the  area. 


Table,  when  the  side 

of  the  polygon  is  1. 

No.  of 

Names. 

Areas,  or 

Radius  of 

sides. 

Multipliers. 

inscribed  circle 

3 

Trigon  or  equil.  A 

0.433013 

0.288675 

4 

Tetragon  or  square 

1.000000 

0.500000 

5 

Pentagon 

1.720477 

0.68S191 

6 

Hexagon 

2.598076 

0.866025 

7 

Heptagon 

3.633912 

1.0i38262 

8 

Octagon 

4.828427 

1.207107 

9 

Nonagon 

6.181824 

1.373739 

10 

Decagon 

7.694209 

1.538842 

11 

Hendecagon 

9.365640 

1.702S44 

13 

Duodecagon 

11.196152 

1.866025 

EXAMPLES. 

1.  The  side  of  a  regular  pentagon  is  12  feet ;  what  is  the 
area? 

Here  12  ^x  1.720477  =  144  x  1.720477  =  247.748688 
feet,  the  area. 

2.  What  is  the  area  of  a  regular  octagon,  the  side  being 
20  feet  ?  ^ns.  1931.3708  ft. 

3.  The  side  of  a  regular  hexagon  is  24  feet ;  what  is  its 
area?  .^ns.  1496.4917  ft. 

4.  What  is  the  area  of  a  regular  heptagon  whose  side  is 
16  yards?  .^ns.  930.2814  yds. 

5.  What  is  the  area  of  a  regular  nonagon  whose  side  is 
36  inches  ?  ^ns.  80 11. 6439  in. 


6. 

from 


iw  many  pieces,  each  4  inches  square,  may  be  cut 
ular  decagon  whose  side  is  12  inchc^s  ? 

^ns.  69.2478  pieces. 


MENSURATION    OF    SUPERFICIES.  83 

PROBLEM  III. 

When  the  area  of  any  regular  polygon  is  given,  to  find 
the  side. 

Rule. — Divide  the  area  by  the  number  in  the  table  cor- 
responding with  the  figure,  and  the  square  root  of  tlie 
quotient  will  be  the  length  of  the  side. 

EXAMPLES. 

1.  The  area  of  a  regular  pentagon  is  4  a  res ;  how  many 
perches  are  contained  in  the  side  ? 

Here  160  X  4  =  640  perches, 

Then  V  (G40  h-  1.7204T7)  =  ^  371.9898  =  19.2870 

perches,  the  length  of  the  side. 

2.  Required  the  length  of  the  side  of  a  regular  hexagon 
containing  one  acre.  ^ns.  7.8475  perches. 

3.  The  area  of  an  octagonal  floor  is  560  feet ;  what  is  the 
length  of  the  side  ?  Ans.  10.7693  feet. 

4.  The  area  of  a  regular  hexagon  is  73.9  feet ;  what  is 
the  side  ?  Ans.  51  feet. 

5.  The  area  of  a  regular  heptagon  is  1356.6  yards;  what 
is  the  length  of  the  side  ?  Ans.  19.3214  yards. 

6.  The  area  of  a  regular  decagon  is  3233.4912  feet ;  what 
is  the  length  of  the  side  ?  Ans.  201  f''et. 


IRREGULAR  FIGURES. 


PROBLEM  L 


§  13.  To  find  the  area  of  an  irregular  right-lined  figure 
of  any  number  of  sides. 

Rule  1. — Divide  the  figure  into  triangles  and  trapeziums, 
and  find  the  areas  of  each  of  them  separately  by  Prob.  I.  p.  67, 
and  Prob.  I.  p.  79 ;  then  add  these  areas  together,  and  their 
sum  will  give  the  area  of  the  whole  figure. 


84 


MENSURATION    OF    SUPERFICIES. 


EXAMPLES. 

1 .  Reqi  lired  the  area  of  the  i  rregular 
right-iined  figure,  A  B  C  D  E  F,  the 
dimensions  of  which  are  as  follows  : 
F  B  =  20.75,  F  C  =  27.48.  E  C  = 
18.5,  B  n  =  14.25,  E  m  =  9.35, 
D  r  =  12.8,  and  A  s  =  8.6  perches 
respectively. 

Here  (F  B  x  A  s)  -^  2  =  (20.75  a 

X  8.6)  -f-  2  =  178.45  -~  2  =  89.225  perches,  the  area  of  the 
triangle,  A  B  F. 

And  (E  C  X  D  r)  -?-  2  =  (18.5  x  12.8)  -^  2  =  236.8  -i- 
2=1 18.4  perches,  the  area  of  the  triangle,  DEC. 

Also,  [(Bn  +  Em)  x  F  C]  ^  2  =  [(14.25  +  9.35)  x 
27.48]-^  2  =  (23.6  x  27.48)  -r-  2=  648.528  -f-  2  =  324.264 
perches,  the  area  of  the  trapezium,  F  B  C  E. 

Whence  324.264  -f  89.225  +  118.4  =  531.889  perches 
=  3  a.  1  r.  11.889  p.,  the  area  of  the  whole  figure. 

2.  Required  the  area  of  an  irregular  hexagon,  like  that  in 
the  last  example,  supposing  the  dimensions  of  the  different 
lines  to  be  the  halves  of  those  before  given. 

Ans.  3  r.  12.9722  p. 

Rule  2. — The  area  of  any  irregular  right-lined  figure  may 
also  be  determined,  by  drawing  perpendiculars  from  all  its 
angles,  to  one  of  its  diagonals,  considered  as  a  base ;  and 
then  adding  the  areas  of  all  the  triangles  and  trapezoids  to- 
gether for  the  content. 


EXAMPLES. 

1.  Required  the  area  of  the  irregular  right-lined  figure, 
A  B  C  D  E  F,  the  dimensions  of  which  are  as  follows  : 
E  n  =  4.54,  E  wi  =  8.26,  E  r  =20.01,  E  s  =  26.22,  E  B 
=  30.15,  D  m  =  10.56,  C  r  =  12.24,  F  n  =  8.56,  and  A  » 
•=  9.26  chains  respectively. 

Here  m  r  =Er  — E  m  =20.01 
—  8.26  =11.75,  n»  =  E«— En  = 
26.22  —  4.54  =  21 .68,  B  r  =  E  B — 
E  r  -=  30.15  —  20.01  =  10.14,  and 
8«=EB  —  Es=  30.15  —  26.22 
<=3.93. 


MENSURATION    OF    SUPERFICIES.  85 

Whence  (F  n  x  E  n)-4-2=(8.56  x  4.54)  -^2  =  38.8624 
-i.  2  =  19.4312  chains,  area  of  the  triangle  E  F  n. 

And  ^Dm  X  Em)  •--  2  =  (10.5G  x  8.26)  -^2=  87.2256 
^  2  =  43.6 128  chains,  area  of  the  triangle,  E  D  m. 

Also  (C  r  X  B  r)  -T-  2  =  (12.24  x  10. 14)  -r-  2  =  124.1 136 
^.  2  =  62.0568  chains,  area  of  the  triangle  C  B  r. 

And  (A  s  X  B  s)  -T-  2  =  (9.26  x  3.93)  -j-  2  =  36.3918  -^ 
2  =  18.1959  chains,  area  of  the  triangle  A  B  s. 

Then  [(D  m  +  C  r)  x  m  r]  -=-  2  =  [(10.56  +  12.24)  X 
11.75]  -r-  2=  (22.8  X  11.75)  -^  2  =  267.9  -~2=  133.95 

chains,  area  of  the  trapezoid  D  C  r  m. 

And  [(F  n  +  A  s)  X  n  s]  --  2  =  [(8.56  +  9.26)  x  21.68] 
-V-  2  =  (17.82  X  21.68)  -^  2  =  386.3376  -i-  2  =  193.1688 
chains,  area  of  the  trapezoid  F  A  «  n. 

Hence  19.4312  +  43.6128  +  62.0568  +  18.1959+133.95 
+  193.1688=  470.4155  chains  =  47  a.  0  r.  6.648  p.,  the 
area  of  the  whole  figure. 

2.  Required  the  area  of  an  irregular  figure,  like  that  in  the 
last  example,  only  doubling  the  dimensions  of  the  diagonal, 
■ind  the  several  perpendiculars,     ^ns.  188  a.  0  r.  26.592  p. 


PROBLEM  II. 

To  find  the  area  of  a  mixtilineal  figure,  or  one  formed  by 
right  lines  and  curves. 

Rule  1. — Take  the  perpendicular  breadths  of  the  figure 
in  several  places,  at  equal  distances  from  each  other,  and 
divide  their  sum  by  their  number,  for  the  mean  breadth  ;  and 
this  quotient,  multiplied  by  the  length,  will  give  nearly  the 
true  area  of  the  figure. 

2d.  Or,  if  greater  accuracy  be  required,  take  half  the  sum 
of  the  two  extreme  breadths,  for  one  of  the  said  breadths, 
and  add  it  to  the  others,  as  before  ;  then  divide  this  sum 
by  the  number  of  parts  in  the  base  (instead  of  by  the  number 
of  breadths),  and  the  result  multiplied  by  the  length  will 
give  the  area,  Avith  a  sufficient  degree  of  correctness  to 
answer  most  questions  of  this  kind  that  can  occur.  When 
the  curved,  or  mixti'ineal,  boundary  meets  the  base,  as  is 
8 


86 


MENSURATION    OF    SUPERFICIES. 


frequently  the  case  in  surveying,  the  area  is  found  by 
dividing  the  sum  of  all  the  breadths  by  the  number  of  parts 
in  the  base,  and  then  multiplying  the  result  by  the  length, 
as  before  ;  observing,  in  each  of  these  cases,  that  the  greater 
the  number  of  parts  into  which  the  base  is  divided,  the 
nearer  will  the  approximation  be  to  the  exact  area. 

It  may  likewise  be  further  remarked,  that  if  the  perpen- 
diculars, or  breadths,  be  not  at  equal  distances  from  each 
other,  the  parts  should  be  computed  separately,  as  so  many 
trapezoids,  and  then  added  together,  for  the  area. 


EXAMPLES. 

1.  The  perpendicular  breadths 
of  the  irregular  mixtilineal  figure, 
A  B  C  D,  at  5  equidistant  places, 
AEGIB,  being  9.2,  10.5,  8.3, 
9.4,  and  10.7  yards,  and  its  length, 
A  B,  30  yards ;  what  is  its  area  ?  ^       -        - 

Here  (9.2  +  10.5  +  8.3  +  9.4  +  10.7)  -^  5  =  48.1  -f-  5 
=  9.62, 

And  9.63  X  30  ss  192.4  yards,  area  by  the  first  part  of  the 
rule. 

Or, 

(AD  +  BC)  ^  2  =  (9.2  +  10.7)  --  2  =  19.9  --  2 
=  9.95. 

Then  (9.95+10.5+8.3  +  9.4)-4- 4=38.15 -T- 4  =9.5375, 
and  9.5375  x  30  =  190.75  yards,  the  area  by  the  second 
part  of  the  rule. 

2.  Required  the  area  of  the  figure, 
A  B  C  D,  of  which  the  part  A  C  is 
a  rectangle,  whose  sides  are  20.i 
and  IO5  chains  respectively;  and 
the  perpendicular  breadths  of  the 
curvilineal  spaces,  reckoning  from 
DC  at  4  equidistant  places,  are  10.2, 
8.7,  10.9,  and  8.  5  chains  respectively. 

Here  (10.2  +  8.7  +  10.9  +  8.5)  -^  5  =38.3  -4-  5  =  7.66, 
And  7.66  x  20.5  =  157.03  chains,  the  area  of  the  curved 

215.25  chains,  the  area  of  the  rect' 


space. 

Then  20.5  x  10.5 
angle. 

Whence  215.25  +  157.03  =  372.28  chains  =  37  a 
36.48  p.,  the  area  of  the  whole  figure. 


Or 


MENSURATION    OF   SUPERFICIES.  87 

3.  The  length  of  an  irregular  mixtilineal  figure  is  47 
chains,  and  Us  breadth,  at  6  equidistant  places,  beginning  at 
the  left  hand  extremity  of  the  base,  5.7,  4.8,  7.5,  5.1,  8.4, 
and  6.5  chains  respectively  ;  what  is  its  area? 

£ns.  29  a.  3  r.  2§  p..  by  the  first  rule. 

4.  The  length  of  an  irregular  mixtilineal  figure,  of  which 
the  curvilinear  boundary  meets  the  base,  is  37^  chains,  and 
its  breadth,  at  7  equidistant  places,  is  4.9,  5.6,  4.5,  8.2,  7.3, 
5.9,  and  8.5  chains  respectively  ;  what  is  its  area  ? 

Ans.  24  a.  0  r.  8.571  p.  by  the  first  rule. 
And  23  a.  3  r.  20  p.  by  the  second  rule. 

5.  The  length  of  an  irregular  field  is  39  rods,  and  its 
breadths,  at  five  equidistant  places,  are  2.4,  2.6,  2.05,  3.65 
3.6  rods  respectively  ;  what  is  the  area  ? 

Ans.  111.54  rods  by  the  first  rule. 

(i.  The  length  of  an  irregular  piece  of  land  being  42 
chains,  and  the  breadths,  at  six  equidistant  points,  being  8.7, 
10.3,  7.1, 8.24,  10.04,  12.2  chains  respectively ;  what  is  the 
area  ?  Ans.  38  a.  2  r.  39.872  p.,  by  the  second  rule. 


THE  CIRCLE. 
PROBLEM  I. 


§  14.  To  find  the  circumference  of  a  circle,  when  the 
diameter  is  given,  or  the  diameter,  when  the  circumference 
is  given. 

Rule. — Multiply  the  diameter  by  3.1416,  and  the  product 
will  be  the  circumference ;  or  divide  the  circumference  by 
3.1416,  or  multiply  the  circumference  by  .31831,  and  the 
result  will  be  the  diameter. 

EXAMPLES. 

1.  What  is  the  circumference  of  the 
circle,  A  C  B  D,  whose  diameter,  A  B,  is  7 
feet  ?  A 

Here  3.14ir)  x  7  =  21.9912  feet,  the  cir- 
cumference. 


88  MENSURATION    OF   SUPERFICIES. 

2.  What  is  the  diameter  of  a  circle  whose  circumference 
is  100  yards? 

Here  100 -j-  3.1416  =  31.831  yards,  the  diameter. 

Or, 
100  X  .31831  ='81.831  yards,  the  diameter  as  before. 

3.  If  the  diameter  of  a  circle  be  17  chains ;  what  is  the 
circumference  ?  ^ns.  53.4072  chains. 

4.  If  the  circumference  of  a  circle  be  354  perches  ;  what 
is  the  diameter  ?  ^ns.  112.6817  perches. 

5.  What  is  the  circumference  of  the  earth,  supposing  its 
diameter  to  be  7935  miles,  which  it  is  very  nearly  ? 

Ans.  24928.590  miles. 

6.  If  the  circumference  of  a  carriage-wheel  be  16  feet  6 
inches ;  what  is  its  diameter  ?  Ans.  5.2521  feet. 

PROBLEM  II. 

To  find  the  area  of  a  circle. 

Rule. — Multiply  the  square  of  the  diameter  by  .7854 ; 
or  the  square  of  the  circumference  by  .07958,  and  the  pro- 
duct in  either  case  will  be  the  area. 

EXAMPLES. 

1.  How  many  square  feet  are  there  in  a  circle  whose 
diameter  is  5  feet  6  inches  ? 

Here  (5.5)^  X  .7854  =  30.25  x  .7854  =  23.75835  sq.  feet. 

2.  Required  the  area  of  a  circle,  the  circumference  of 
which  is  9i  yards. 

Here  (9.2)  ^  x  .07958  =  84.64  X  .Q7958  =  6.7356512 
sq.  yds. 

3.  How  many  square  yards  are  there  in  a  circle  whose 
radms  is  15|  feet?  Ans.  81.1798  sq.  yards. 

4.  How  many  square  feet  are  there  in  a  circle  whose 
circumference  is  10?  yards  ?  Ans.  82.7681  sq.  feet. 

5.  The  diameter  of  a  circle  is  16  chains  ;  how  many  acres 
does  it  contain  ?  Ans.  20  a.  0  r.  16.99:54  p. 

6.  What  is  the  value  of  a  circular  garden  whose  diameter 
is  6  perches,  at  the  rate  of  75  cents  per  square  yard  ? 

Ans.  $641,475. 


MENSURATION    OF    SUPERFICIES.  89 


PROBLEM  III. 

The  area  of  a  circle  being  given,  to  find  the  diameter  or 
circumference. 

Rule. — ^Divide  the  area  by  .7854,  and  the  square  root  of 
the  quotient  will  be  the  diameter.  Or,  divide  the  area  by 
.07958,  and  the  square  root  of  the  quotient  will  be  the  cir- 
cumference. 

EXAMPLES. 

1.  The  area  of  a  circle  is  5  acres,  3  roods,  and  2G  perches  ; 
what  is  the  diameter  ? 

Here  5  a.  3  r.  20  p.  =  946  perches  ;  and  v/(946  -^  .7854) 
=  v/  1204.48179271  =  34.7056  perches,  the  diameter. 

2.  The  area  of  a  circle  being  2  acres,  3  roods,  and  12 
perches,  what  is  the  circumference  ? 

Here  2  a.  3  r.  12  p.  =  452  perches  ;  and  ^  (452-f-  .07958) 
=  y/  5679.09  =  75.3637  perches,  the  circumference. 

3.  The  area  of  a  circle  is  5028f  square  yards  ;  what  is  its 
diameter  ?  Ans.  80.0160  yds. 

4.  The  area  of  a  circular  garden  being  1  acre,  what  is  the 
length  of  a  stone  wall  which  will  enclose  it  ? 

Ans.  44.8392  p. 

5.  It  is  required  to  find  the  radius  of  a  circle  whose  area 
is  an  acre.  Ans.  39.2507  yds. 

6.  The  value  of  a  circular  piece  of  ground,  at  $4  per 
square  perch,  is  $40.50.  How  many  dollars  will  encircle  it 
if  the  diameter  of  a  dollar  be  I5  inches? 

Ans.  $1595.3916. 

PROBLEM  IV. 

To  find  the  area  of  a  circular  ring,  or  the  space  included 
between  two  concentric  circles. 

Rule. — Find  the  areas  of  the  two  circles  separately. 
Then  the  difference  of  these  areas  will  be  the  area  of  the 
r*^g. 

Or,  multiply  the  sum  of  the  diameters  by  their  difference. 

8* 


90  MENSURATION    OF   SUPERFICIES. 

and  this  product  again  by  .7854,  and  it  will  give  the  area 
required. 


EXAMPLES. 

1.  The  diameters  of  the  two  circles  y^         n. 

are,  A  B  20,  and  D  C  12  yards ;  re-       /^ ^ ^X 

quired  the  area  of  the  ring.  If  \    \ 

Here  400  x  .7854  =  314.16,  area  of  aMc dJ— Jb 

the  outer  circle.  y    V  J    j 

144  X  .7854  =  1 13.0976,  area  of  the      \   ^^ — ^ y 
inner  circle.  ^^____-^ 

And  314.16—113.0976  =  201.0624  yards,  area  of  the 
ring. 

Or,  20  +  12  =  32,  sum  of  diameters. 

20  —  12  =    8,  difference  of  diameters. 

And  32  X  8  X  .7854  =  201.0024  yards,  area  as  before. 

2.  What  is  the  area  of  a  circular  ring,  the  diameters  of 
the  concentric  circles  being  20  and  30  feet  ?  Ans.  392.7  ft. 

3.  The  diameters  of  two  concentric  circles  are  8  and  12 
yards  ;  what  is  the  area  of  the  ring  contained  between  their 
circumferences  ?  Ans.  62.832  yds. 

4.  The  diameters  of  two  circles  are  21|  and  9^  feet;  re- 
quired the  area  of  the  ring.  Ans.  300.6009  ft. 

5.  The  diameter  of  the  inner  circle  is  6,  and  the  outer  10 
chains  ;  what  is  the  area  of  the  ring  ?     Ans.  50.2656  chs. 

6.  The  area  of  the  outer  circle  contains  100  acres,  and 
the  diameter  of  the  inner  is  equal  to  |  of  the  diameter  of  the 
greater ;  what  is  the  area  of  the  ring  ?       Ans.  555.55  chs. 


PROBLEM  V. 

The  diameter  or  circumference  of  a  circle  being  given,  to 
find  the  side  of  an  equivalent  square. 

Rule. — Multiply  the  diameter  by  .8862, "or  the  circum- 
ference by  .2821,  the  product  in  either  case  will  be  the  side 
of  an  equivalent  square. 


MENSURATION    OF   SUPEKFICIES.  91 

EXAMPLES. 

1.  The  diameter  of  a  circle  is  200  yards  ;  what  is  the  side 
of  a  square  of  equal  area  ? 

Here  200  X  .8862  =  177.24  yards. 

2.  The  circumference  of  a  circle  is  316  yards ;  what  is 
he  side  of  a  square  of  equal  area  ? 

Here  310.  x  .2821  =81).  1436  yards. 

3.  The  diameter  of  a  circle  is  1 142  feet ;  what  is  the  side 
of  a  square  of  equal  area?  .^ns.  1012.0-104  ft. 

4.  The  circumference  of  a  circle  is  18.8  chains  ;  what  is 
the  side  of  a  square  of  equal  area  ?  ^ns.  5.3034  chs. 

5.  The  diameter  of  a  circular  fish-pond  is  22  perches ; 
what  would  be  the  side  of  a  square  fish-pond  of  an  equal 
area  ?  Jns.  19.4964  p. 

6.  The  circumference  of  a  circular  walk  is  64  rods  ;  what 
Is  the  side  of  a  square  containing  the  same  area  ? 

^7is.  18.0544  r. 


PROBLEM  VI. 

The  diameter  or  circumference  of  a  circle  being  given,  19 
find  the  side  of  the  inscribed  square. 

RuijE. — Multiply  the  diameter  by  .7071,  or  the  circum- 
ference by  .2251,  and  the  product  in  either  case  will  be  the 
side  of  the  inscribed  square. 

EXAMPLES. 

1 .  The  diameter,  A  B,  of  a  circle  is 
614  feet ;  what  is  the  side,  A  C,  of  the 
inscribed  square  ? 

Here  614  x  .7071  =  434.1594  feet 
=  AC. 

2.  The  circumference  of  a  circle  is 
1804  feet;  what  is  the  side  of  the  in- 
scribed square  ?         ^ns.  406.0804  ft. 

3.  The  diameter  of  a  circle  is  239  feet ;  what  is  the  side 
of  the  inscribed  square  ?  Jins.  168.9909  fl. 


92  MENSURATION    OF   SUPERFICIES. 

4.  The  circumference  of  a  circle  is  98  chains  ;  what  is 
the  side  of  the  inscribed  square  ?  Ans.  22.059S  chs. 

5.  The  diameter  of  a  circle  is  65  rods  ;  what  is  the  side 
of  the  inscribed  square  ?  ^ns.  45.9015  r. 

6.  The  circumference  of  a  circular  walk  is  721  perches  ; 
wJiat  is  the  side  of  an  inscribed  square  ? 

Ans.  162.2971  p. 


PROBLEM  VII. 

To  find  the  diameter  of  a  circle  equal  in  area  to  any  given 
superficies. 

Rule. — Divide  the  area  by  .7854,  and  the  square  root  of 
the  quotient  will  be  the  diameter. 

EXAMPLES. 

1.  The  length  and  breadth  of  a  rectangle  are  24  and  10 
chains  ;  what  is  the  diameter  of  a  circle  which  contains  the 
same  area? 

Here  24  x  16  =  384,  the  area  of  the  rectangle. 
Then  ^(384  -^  .7854)  =  v/488.9228=  22.1116  chains, 
the  diameter. 

2.  The  side  of  a  square  is  16  perches  ;  what  is  the  diame- 
ter of  a  circle  containing  the  same  area  1 

Ans.  18.054  p. 

3.  The  base  and  perpendicular  of  a  right-angled  triangle 
are  16  and  20  feet ;  what  will  be  the  diameter  of  a  circle 
which  contains  the  same  area  ?  Ans.  14.2729  ft. 

4.  The  three  sides  of  a  scalene  triangle  are  14, 18,  and  24 
yards ;  what  is  the  diameter  of  a  circle  containing  the  sam 
area?  .^ns.  12.0267  yds, 

5.  The  three  sides  of  a  triangle  are  18,  20,  and  2(5  feet ; 
what  IS  the  diameter  of  a  circle  containing  three  times  as 
much?  .^n*.  20.1919  ft. 

6.  The  length  and  breadth  of  a  parallelogram  are  :^2  and 
18  yards :  what  is  the  diameter  of  a  circle  that  contains  the 
same  area  ?  Ans.  27.0810  \  ds. 


MENSURATION    OF   SUPERFICIES.  93 


PROBLEM  VIII. 

The  diameter  of  a  circle  being  given,  to  find  another  con- 
taining a  proportionate  quantity. 

■  Rule. — Multiply  the  square  of  the  given  diameter  by  tho 
given  proportion,  and  the  square  root  of  the  product  will  be 
the  diameter  required. 


EXAMPLES. 

1.  The  diameter  of  a  circle  is  24  chains ;  what  is  the 
diameter  of  one  containing  one-fourth  of  the  area  ? 

Here  ^{24^  X  ?)  ==  v/(576  x  ^)  =  v/144  =  12  chains. 

2.  The  diameter  of  a  circle  is  16  perches  ;  what  is  the 
diameter  of  one  containing  nine  times  as  much  ? 

^ns,  48  p. 

3.  The  diameter  of  a  circle  is  36  yards;  what  is  the 
diameter  of  one  containing  four  limes  as  much  ? 

^ns.  72  yds. 

4.  The  diameter  of  a  circle  is  81  feet ;  what  is  the  diame- 
ter of  one  containing  five  times  as  much  ? 

^7is.  181.1215  ft. 

5.  A  gentleman  has  a  circular  grass-plat  in  his  yard,  the 
diameter  of  which  is  25  yards ;  required  the  length  of  the  string 
that  would  describe  a  circle  to  contain  sixteen  times  as  much. 

^ns.  50  yds. 

6.  The  diameter  of  a  circle  is  9  rods  ;  what  is  the  diame- 
ter of  one  containing  six  times  as  much?    ^ns.  22.0454  r. 


PROBLEM  IX. 

To  find  the  length  of  any  arc  of  a  circle. 

Rule.  I . — From  eight  times  the  chord  of  half  the  arc,  si  "b- 
tract  the  chord  of  the  whole  arc,  and  one-third  of  the  '"?- 
mainder  will  U;  the  length  of  the  arc  nearly. 


94  MENSITRATION    OF    SUPERFICIES. 

The  chord  of  the  whole  arc,  or  sim- 
ply the  chord,  is  a  right  line  which  joins 
the  extremities  of  an  arc.  Thus  in  the 
figure,  D  C  is  the  chord  of  the  whole 
arc,  D  F  C  ;  D  F  the  chord  of  half  the 
arc,  D  F  C  ;  and  D  E,  or  E  C,  half  the 
chord  of  the  arc  D  F  C.  The  height  of 
an  arc,  called  its  versed  sine,  is  that  part 
of  the  diameter  contained  between  the 
middle  of  the  chord  and  the  arc.  Thus,  in  the  case  of  the 
arc  D  F  C,  the  versed  sine,  or  the  height  of  the  arc,  is  the 
line  E  F. 

The  chord  of  half  the  arc  may  be  found  by  adding  together 
the  square  of  half  the  chord  and  the  square  of  the  versed 
sine,  and  extracting  the  square  root  of  the  sum. 

Or,  by  taking  the  square  root  of  the  product  of  the  diame- 
ter and  the  versed  sine. 

Half  the  chord  of  the  whole  arc  may  be  found  by  sub- 
tracting the  versed  sine  from  the  diameter,  multiplying  the 
remainder  by  the  versed  sine,  and  taking  the  square  root  of 
the  product.  By  doubling  this  we  get  the  chord  of  the 
whole  arc. 

To  find  the  versed  §ine,  or  height  of  the  arc,  subtract  the 
square  of  the  chord  from  the  square  of  the  diameter,  and  ex- 
tract the  square  root  of  the  remainder ;  subtract  this  root  from 
the  diameter,  and  one-half  of  the  remainder  Avill  be  the  versed 
sine. 

Or,  from  the  square  of  the  chord  of  half  the  arc,  subtract 
the  square  of  half  the  chord  of  the  arc,  and  the  square  root 
of  the  remainder  will  give  the  versed  sine. 

Again,  to  obtain  the  versed  sine,  divide  the  square  of  the 
chord  of  half  the  arc  by  the  diameter. 

The  diameter  may  be  known  by  adding  together  the  square 
of  the  versed  sine  and  the  square  of  half  the  chord  of  the  arc, 
and  dividing  the  sum  by  the  versed  sine.  The  diameter 
may  hkewise  be  obtained  by  dividing  the  square  of  the  chord 
of  half  the  arc  by  the  versed  sine. 

It  may  here  likewise  be  observed,  as  another  rule  for  the 
Fame  purpose,  that  if  the  number  of  degrees  in  the  arc  be 
multiplied  by  radius,  and  that  product  again  by  .01745,  the 
result  will  give  the  length  of  the  arc  nearly. 


MENSURATION    OF    SUPERFICIES. 


95 


EXAMPLES. 

1.  The  chord  A  B  of  the  whole  arc 
A  C  B,  is  48,74  feet,  and  the  chord  A  C 
of  half  the  arc  30.25  feet ;  what  is  the 
length  of  the  arc  ? 

Here  [  (30.25  x  8)  — •  48.74)  ]  ^  3  = 
(242  —  48.74)  -4-  3  =  193.26  -=-  3  = 
64.42  ft.  the  length  of  the  arc  A  C  B  nearly. 

2.  If  the  chord  A  B,  of  the  arc  A  C  B,  he  30  yards,  and 
the  versed  sine,  or  height  C  D,  8  yards,  what  is  the  length 
of  the  arc  ? 

Here  ^/  (15«  +  S^J  =  y/  (225  +  64)  =  ^Z  289  =  17  = 
A  C,  the  chord  of  half  the  arc. 

And  [(17  X  8)  —  30)]  -h  3=(136  — 30)  -f-  3=106 -4-  3 
=  35?  yards,  the  length  of  the  arc  nearly. 

.3.  If  the  versed  sine,  or  height  of  half  the  arc,  be  4  feet, 
and  the  diameter  of  the  circle  30  feet,  what  i.*;  the  length  of 
the  arc  ? 

Here  ^(30  x  4)  =  v/120  =  10.95445  =.  A  C,  the  chord 
of  half  the  arc. 

And  v/  [  (30  —  4)  X  4]  X  2  =  ^{26  x  4)  x  2  =  v/(104) 
X  2=  10.19803  X  2 =20.39606  =  A  B,  the  chord  of  the 
whole  arc. 

Then  [(10.95445  x  8)  —  20.39606]  -^  3  =  (87.6356 
—  20.39606)  -T-  3=  67.23954  -r-  3  =  22.41318  feet,  the 
length  of  the  arc  nearly. 

4.  Required  the  length  of  an  arc  of  30  degrees,  the  radius 
of  the  circle  being  14  feet. 

(30  X  14  X  .01745)  =  7.329  feet,  the  length  of  the  arc. 

5.  The  chord  of  the  whole  arc  is  50|  yards,  and  the  chord 
of  half  the  arc  is  33^  yards ;  required  the  length  of  the  arc. 

^ns.  71.6  yards. 

6.  The  length  of  the  chord  of  the  whole  arc  is  36|  feet, 
and  the  length  of  the  chord  of  half  the  arc  is  231  feet ;  what 
is  the  length  of  the  arc  ?  ^ns.  49.6166  feet. 

7.  The  chord.of  the  whole  arc  is  48$  feet,  and  its  versed 
sine,  or  height  of  the  arc,  18|  feet ;  what  is  the  length  of  the 
arc?  ^n«.  64.7667  feel. 


96  MENSURATION    OF   SUPERFICIES. 

8.  If  the  versed  sine  or  height  of  the  arc  be  2  yards,  and 
the  diameter  of  the  circle  36  yards  ;  what  is  the  length  of 
the  arc  ?  Ans.  17. 1299  yards. 

9.  If  the  chord  of  the  whole  arc  be  16  chains,  and  the 
radius  of  the  circle  10  chains  ;  what  is  the  length  of  the  arc  ? 

Ans.  18.518  chains. 

10.  The  chord  of  half  the  arc  is  10  perches,  and  the 
diameter  of  the  circle  16§  perches ;  what  is  the  length  of 
the  arc?  Ans.  21?  perches. 

11.  Required  the  length  of  an  arc  of  57°  17'  445",  the 
diameter  of  the  circle  being  50  feet. 

Ans.  25  feet,  which  equals  the  radius. 

12.  Required  the  length  of  a  degree  of  a  great  circle  of 
the  earth,  supposing  its  circumference  to  be  25000  miles. 

Ans.  695  miles  nearly. 

Rule  2. — Let  d  =  the  diameter  C  E  of  the  circle,  and 
V  =  the  versed  sine  or  height  C  D  of  half  the  arc,  then  will 
the  length  of  the  arc  be  expressed  by  the  following  series : 


6x7rf       '8x9(/ 
Where  A,  B,  C,  »fec.  represent  the  terms  immediately  pre- 
ceding those  in  which  they  first  occur. 

Which  rule  is  to  be  used,  by  substituting  in  the  above 
series  the  numerical  values  of  the  given  parts,  in  the  place 
of  the  letters  by  which  they  are  denoted,  and  then  finding  the 
sum  of  such  a  number  of  its  terms  as  may  be  thought  suffi- 
cient for  determining  the  length  of  the  arc,  to  a  degree  of 
accuracy  required. 


c 


examples; 


1.  Required  the  length  of  the  arc, 
A  C  B,  whose  versed  sine  or  height, 
DC,  is  5  feet,  and  the  diameter,  CE,  of 
vhe  circle  25  feet. 

Here  d  —.  25,  and  u  =  5. 


MENSURATION    OF    SUPERFICIES.  97 

Then  2  ^  rfu  =  3  ^(25  x  5)  =  2  v/ 125  =  22.3606798  ==  A 

_Ji_  X  A  =  3-A^  X  22.3606798  =   .7453559=  B 
2xS  d  6  X  25 

'^^      X  B  =  ?^  ^  L  X   .7453559  =   .0670820 = C 


4  X  5  rf      20  X  25 
^' "   X  C  =  ^l"  ^?  X   0G70820  =   .0079859  =  D 


6x7  d  42  X  25 

7a  y  49  V  5 

X  D  =  — ^-^r;:  X      .0079859  =      .0010869 =E 


Qxiid^  72  X  25  ^,  _.-  looionr^ 

The  sum  =  23.1821905 

feet,  =  the  length  of  the  arc,  A  C  B. 

2.  Required  the  length  of  the  arc,  A  C  B,  whose  versed 
sine  or  height,  DC,  is  2  feet,  and  the  diameter,  C  E,  of  the 
circle  52  feet.  ^ns.  20.5291  feet. 

3.  Required  the  length  of  the  arc  whose  versed  sine  is  9 
yards,  and  the  diameter  of  the  circle  100  yards. 

^ns.  60.9385  yards. 

4.  It  is  required  to  find  the  length  of  the  arc  whose  chord 
is  16  perches,  and  its  height  4  perches. 

Jlns.  18.5459  perches. 

5.  It  is  required  to  find  the  length  of  the  arc  whose  height 
is  6  inches,  and  the  chord  of  half  the  arc  1  foot. 

^ns.  2.0943  feet. 

6.  It  is  required  to  find  the  length  of  the  arc  whose  chord 
is  48  chains,  and  the  radius  of  the  circle  25  chains. 

Ans.  64.3501  chains. 


PROBLEM  X. 
To  find  the  area  of  a  sector  of  a  circle. 

Role  1. — Multiply  the  radius  or  half  the  diameter  of  the 
circle  by  half  the  length  of  the  arc  of  the  sector,  as  found  by 
the  last  Problem,  and  the  product  will  be  the  area. 

To  this  we  may  add,  that  as  360°  is  to  the  number  of  de- 
grees in  the  arc  of  the  sector,  so  is  the  area  of  the  circle,  to 
the  area  of  the  sector. 

9 


c 


MENSURATION    OF    SUPERFICIES. 


EXAMPLES. 

1  The  chord,  A  B,  of  the  whole  arc, 
A  C  B,  is  24  feet,  and  the  chord,  A  C, 
of  half  the  arc  l:i  feet ;  what  is  the  area 
of  the  sector,  OBCAO? 

Herev/(13^  — 13^)=>/(169  — 144)     \  ;" 

=  ^  25  =  5=C  D,  the  versed  sine.  \  I         /' 

And   (13*  -i-  5)  =169  -r-  5=33.8         ^---..L.--'' 
=  C  E,  the  diameter.  ^ 

[f  13  X  8)  —  24]  ^  3  =  (104  —  24) -=- 3  =  80 -=- 3  =  3G§ 
=  tne  length  of  arc  A  C  B, 

Then  13^  x  16.9  =  225?  feet,  the  area  of  the  sector. 

2.  Required  the  area  of  the  sector,  the  arc  of  which  is  30°, 
and  the  diameter  3  yards. 

Here  .7854  x  9  =  7.0686  =  the  area  of  the  circle. 
As  360°  :  30°  : :  7.0686  :  .58905  yards,  the  area  of  the 
sector. 

3.  The  chord  of  the  whole  arc  is  8  yards,  and  its  height 
3  yards  ;  what  is  the  area  of  the  sector  ? 

Ans.  22.2222  yards. 

4.  What  is  the  area  of  a  sector  whose  chord  is  18|  chains, 
and  the  diameter  of  the  circle  20  chains  ? 

Ans.  118.954  chains. 

5.  Required  the  area  of  the  sector  whose  height  is  4 
perches,  and  the  radius  of  the  circle  8  perches. 

Ans.  66.8581  perches. 

6.  Required  the  area  of  the  sector  whose  arc  is  17°  15', 
and  the  diameter  of  the  circle  19  feet. 

Ans.  13.5857  feet. 


Rule  2. — Let  d  =  the  diameter  C  E  of  the  circle,  and 
V  =  the  versed  sine  or  height,  C  D,  of  half  the  arc,  then 
will  the  area  of  the  sector  be  expressed  by  the  following 
series. 

Area  O  AC  BO  =  d  rf  v/ ^u  +  s-^,  A  + -^^^.B  + 


2x3(/       '  4x5t/ 


5"  V  7^v 

6x7rf  8x9tZ 


MENSURATION    OF    SUPERFICIES. 


99 


When  A,  B,  C,  &c.,  represent  the  terms  immediately 
preceding  those  where  they  first  occur.  This  Rule  is  to  be 
used  in  the  same  manner  as  directed  in  regard  to  Rule  2, 
Problem  IX.  p.  96. 


EXAMPLES. 


1.  Required  the  area  of  the  sector, 
0  A  C  B  O,  whose  versed  sine,  or  height, 
C  D,  is  7  feet,  and  the  diameter,  C  E,  of 
the  circle  28  feet. 


Here  rf  =  28,  and  v  =  7.  ^ 

Then  ^f/^/rf?;=14^/(28x7)=14^/196=  196.000000  =  A 

7 


flx'Sd 

S'v 

4x5rf 

S'u 

6X7(1 

7»v 

8x9(/ 

9«u 

10  X  lid 

IPu 

XA  = 

6x28^ 

196           = 

8.160666  =  B 

XB  = 

9x7 

20  X  28 

8.166666  = 

.918750  =  0 

XC  = 

25X7 
42  X  28  ^ 

.918750  = 

.136718  =  D 

XD  = 

49x7 
72  X  28  ^ 

.136718  = 

.023261  =  E 

XE  = 

81x7 
110  X  28 

.023261 = 

.004282  =  F 

XF  = 

121x7 
l!^6  ^28 

.004282  = 

.0O0a30  =  G 

12  x  13  d 

The  sum  =  205.250507 
feet  =  the  area  of  the  sector  O  A  C  B  O. 

2.  Required  the  area  of  the  sector  O  A  C  B  O,  whose 
versed  sine,  C  D,  is  2  yards,  and  the  diameter,  C  E,  of  th 
circle  52  yards.  ^ns.  266.8787  yds. 

3.  Required  the  area  of  the  sector,  the  radius  of  the  cir- 
cle being  10  feet,  and  the  chord  of  the  arc  12  feet. 

£ns.  64.3501  ft. 

4.  Required  the  area  of  the  sector  of  a  circle  whose  diame- 
ter is  20  perches,  and  the  chcrd  of  its  arc  16  perches 

Jlns  92.7295  p 


100  MENSURATION    OF    SUPERFICIES. 

5.  Required  the  area  of  the  sector  of  a  circle  whose  versed 
sine  is  9  feet,  and  the  chord  of  half  the  arc  30  feet. 

^ns.  1523.4632  ft. 

6.  Required  the  area  of  the  sector  of  a  circle  whose  chord 
is  48  feet,  and  versed  sine  18  feet.  ^ns.  804.370  ft. 


PROBLEM  XT. 
To  find  the  area  of  a  segment  of  a  circle. 

Rule  1. — Find  the  area  of  the  sector,  having  the  same 
arc  as  the  segment,  by  the  last  problem. 

Also  find  the  area  of  the  triangle  formed  by  the  chord  of 
the  segment,  and  the  two  radii  of  the  sector,  by  Problem  I. 
page  67. 

The  sum,  or  difference  of  these  areas,  according  as  the 
segment  is  greater  or  less  than  a  semicircle,  will  be  the  area 
of  the  segment  required. 

EXAMPLES. 

1 .  The  chord  A  B  is  24  feet,  and  the 
versed  sine  or  height  C  D,  of  half  the 
arc  A  C  B,  is  5  feet ;  what  is  the  area 
of  the  segment  A  B  C  A  ? 

Here  ^Z  (A  D^  +  C  D«)  =  v/  (12''  + 
5»)  =  ^/  (144  +  25)  =  ^/  169  =  13  = 
A  C,  the  chord  of  half  the  arc. 

Again,  (A  C  -^  C  D)  =  (169  -i-  5)  =  33.8  =  C  E,  the 
diameter;  therefore  16.9  =C  O,  the  radius. 

Then  [  (13  X  8)  — 24]  ^  3  =  26|,  the  length  of  the  arc 
A  C  B. 

And  (131  X  16.9)  =225.3333  feet,  the  area  of  the  sec- 
tor O  A  C  B  O. 

But  (C  O  —  C  D)  =  (10.9  —  5)  =  11.9  =  perpendicular 
OD. 

Whence  (11.9  x  12)  =  142.8,  area  of  the  A  A  O  B. 

And,  consequently,  (225.3333—142.8)  =82.5333  feet 
area  of  the  segment  A  B  C  A. 

3.  What  is  the  area  of  a  segment  of  a  circle  whose  arc  is 
00°,  and  the  diameter  of  the  circle  10  feet  ? 

Here  .7854  x  100  =  78.54,  area  of  the  whole  circle. 


MENSURATION    OF    SUPEUFTCIES.  101 

Then,  as  360°  :  60°  : :  78.54  :  13.09,  area  of  the  sector 
<>  A  C  B  O. 

And  since  the  chord  A  B  (to  an  arc  of  60°)  is  =  radius 
O  A  or  O  C,  which  =  5,  then  by  Problem  II.,  page  82,  the 
area  of  the  A  A  O  B  =  10.8253. 

Whence  (13.09  —  10.8253)  =  2.2647  feet,  the  area  of 
the  segment  required. 

3.  Required  the  area  of  the  segment  of  a  circle,  whose 
versed  sine,  or  height,  of  half  the  arc  is  5  yards,  and  the 
diameter  of  the  circle  20  yards.  Ans.  61.1045  yds. 

4.  Required  the  area  of  the  segment  of  a  circle,  whose 
chord  is  16  chains,  and  diameter  of  the  circle  16|  chains. 

Ans.  70.2222  chs. 

5.  What  is  the  area  of  the  segment  of  a  circle,  whose  arc 
is  a  quadrant,  the  diameter  being  24  perches  ? 

^«s.  41.0976  p. 

6.  Required  the  area  of  a  segment  of  a  circle,  whose 
chord  is  18.9  feet,  and  height  2.4  feet.       Ans.  30.601  ft. 

Rule  2. — Let  d  equal  the  diameter  C  E  of  the  circle,  and 
V  equal  the  versed  sine  C  D,  of  half  the  arc,  or  height  of  tne 
segment ;  then  will  the  area  of  the  segment  be  expressed  by 
the  following  series. 

Area  of  the  segment  A  C  B  A  = 
4  V  y/d  V         Sv  5v  S  x7v         5x9  v 

3       ~'2xVd^~  4xYd^~  exlT^^"  8x  iid^- 

Where  A,  B,  C,  &c.,  are  the  first,  second,  third,  &c.,  terms 
of  the  series. 

This  rule  is  to  be  used  in  a  manner  simila^  to  that  directed 
in  regard  to  Rule  2d,  Problem  IX.,  page  96. 

EXAMPLES. 

1.  Required  the  area  of  the  segment  /<?n — ~"\ 

A  B  C  A  of  a  circle,  whose  chord  A  B     4^_ _\^ 

is  16  feet,  and  the    diameter  C  E   20  /' X.      »  ^\ 

feet.  I         \L^      ) 

Here^/(CE^  — AB^)=v/(20«— 16^)  \            1°           / 

=  ^/ (400  — 256)  =  v/144  =  12  ;  then  \          !          / 

(20  —  12)  -T-  2  =  4  =  C  D,  the  versed  >  ,_  !^_..,-'' 

sine.  ^ 

9» 


102  J^fiNSURATION    OF   SUPERFICIES,   s 

*vy/dv  _16^(20X  4)  ^      16^/80  ^           ,    ^^ ^^^^gg  ^  ^ 
3  3  3  -^ 

—  ^^      xA=  — -l^ii.  X  4-47.702783  =  — 2.8621 67  =  B 

2x5'^  10x20 

X  B=  —  _A2<i_  X  —  2.862167  =  —  .102220  =  C 


5v 


4  X  7  d  28  X  20 

liilJ?  X  C=  —  ^L2< A  X  —  .102220  =  —  .007950  =  D 
6  X  9  d                     64  X  20 

5  X  9  f       T^^  _  45  X  4    ^_  .007950=  — .000813  =  B 
8Xlld                    88X20 

'^X^^'^yT^ 77X4         _  .000813=  — .000096  =  F 

10  X  13  d  130X20  . 

The  sum  of  B  C  D  E  F=—  2.973246 


The  difference  =        44.729537,  ft. 

=  the  area  of  the  segment  A  B  C  A. 

2.  Required  the  area  of  the  segment  of  a  circle  whose 
height  is  34  feet,  and  the  radius  2G  feet.  £ns.  957.9609  ft. 

8.  Required  the  area  of  the  segment  of  a  circle  whose 
versed  sine  is  3  yards,  and  the  diameter  60  yards. 

^ns.  52.8533  yds. 

4.  Required  the  area  of  the  segment  of  a  circle  whose 
height  is  4  feet,  and  the  radius  50  feet. 

^ns.  105.3773  ft. 

5.  Required  the  area  of  the  segment  of  a  circle  whohe 
chord  is  20  chains,  and  the  diameter  52  chains. 

^ns.  26.8787  chs. 

6.  Required  the  area  of  the  segment  of  a  circle  whose 
height  is  2  feet,  and  chord  12  feet.  ^ns.  16.35  ft. 

Rule  3. — 1.  From  seven  times  the  diameter  subtract  five 
times  the  versed  sine ;  multiply  the  remainder  by  seven 
times  the  versed  sine,  and  extract  the  square  root  of  the 
product. 

2.  Multiply  the  diameter  by  the  versed  sine,  and  extract 
the  square  root  of  the  product. 

3.  To  the  first  root  add  four  thirds  of  the  second,  and 
multiply  the  sum  by  four  twenty-fifths  of  the  versed  sine, 
and  the  product  will  be  the  area  of  the  segment  nearly. 


MENSURATION    OF    SUPERFICIES.  103 


EXAMPLES. 

1.  Required  the  area  of  the  segment, 

A  B  C  A,  of  a  circle  whose  versed  sine,  y-;^^ — \ 

or  height,  C  D,  is  4  feet,  and  the  diame-  a^__ _\ti 

ter,  C  E,  20  feet.  /  \     ^  ^^\ 

Herev/{[(20x7)-(4x5)]x(4x7)}  I       ^<^       \ 

=  v/[(140— 20)  X  28]  =  ^(120x28)  \  i  / 

=  v/  3360  =  57.96550  the  first  root.  \,         !  / 

And  ^/  (20  X  4)  =  v'SO  =  8.94427,  ^^-.1..-''' 

the  second  root.  ^ 

Then  57.96550  +  (8.94427  X  |)= (57.96550+ 11. 92569) 

=  69.89119. 

Whence  69.89119  x  (4  X  ^) =(69.891 19 x. 64) =44.73036 
feet,  the  area  of  the  segment  A  B  C  A. 

2.  What  is  the  area  of  a  segment  of  a  circle  whose  versed 
sine,  or  height,  is  4  feet,  and  the  diameter  of  the  circle  is 
40  feet  ?  Ans.  65.4005  feet. 

3.  Required  the  area  of  the  segment  of  a  circle  whosa 
versed  sine  is  4  yards,  and  the  diameter  100  yards. 

Ans.  105.3773  yards. 

4.  Required  the  area  of  the  segment  of  a  circle  whose 
height  is  3  yards,  and  the  diameter  60  yards. 

Ans.  52.8533  yards. 

5.  What  is  the  area  of  a  segment  of  a  circle  whose  height, 
or  versed  sine,  is  5  feet,  and  the  diameter  of  the  circle  25 
feet?  ^ns.  69.8911  ft. 

6.  What  is  the  area  of  the  segment  of  a  circle  whose 
height,  or  versed  sine,  is  3  feet,  and  the  diameter  of  the 
circle  8  feet  ?  Ans.  17.2198  ft. 


Rule  4. — To  two-thirds  of  the  product  of  the  chord  and 
versed  sine  of  the  segment,  add  the  cube  of  the  versed  sme 
divided  by  twice  the  chord,  and  the  sum  will  give  the  area  of 
the  segment  nearly. 


104  MENSURATION    OF    SUPERFICIES. 


EXAMPLES. 

1.  What  is  the  area  of  the  segment,  y<^ — \^ 
ABC  A,  of  a  circle  whose  versed  sine,  -^^^    i       ^"^ 
C  D,  is  4  feet,  and  the  chord,  A  B,  is  16  /\.     f  y^\ 
feet?  t       ^^       ! 

1  lO  I 

IIere[(16x4)x|]  +  [4'-5-(16x2)]     \  \ 

=  (64x^)  +  (64-32)  =  (i^  +  2) 

=  (42§  +  2)  =  44§  feet,  the  area  of  the  segment  A  B  C  A. 

2.  What  is  the  area  of  the  segment  of  a  circle  whose 
versed  sine  is  2  feet,  and  the  chord  20  feet  ? 

Ans.  26.8§  ft. 

8.  Required  the  area  of  the  segment  of  a  circle  whose 
height  is  2  feet,  and  the  chord  12  feet.  Arts.  I65  ft. 

4.  Required  the  area  of  the  segment  of  a  circle  whose 
versed  sine  is  15  feet,  and  the  chord  40  feet. 

Ans.  442.1875  ft. 

5.  Required  the  area  of  the  segment  of  a  circle  whose 
versed  sine  is  2  feet,  and  the  chord  7  feet.      Ans.  ^\\  ft. 

6.  Required  the  area  of  the  segment  of  a  circle  whose 
versed  sine  is  18  feet,  and  the  chord  48  feet. 

Ans.  6361  ft. 

Rule  5. — Divide  the  versed  sine,  or  height  of  the  seg- 
ment by  the  diameter  of  the  circle,  and  find  the  quotient  in 
the  table  of  verged  sines,  p.  295. 

Then  multiply  the  tabular  area  on  the  right  hand  of  the 
versed  sine  so  found,  (which  is  the  tabular  segment,)  by  the 
square  of  the  diameter,  and  the  product  will  be  the  area. 

When  the  quotient  arising  from  dividing  the  versed  sin 
by  the  diameter  has  a  remainder,  or  fraction,  after  the  third 
place  of  decimals,  subtract  the  tabular  area,  answering  to  the 
first  three  figures,  from  the  next  following  area  ;  then  if  the 
remainder  be  multiplied  by  the  fractional  part,  and  the  result 
be  added  to  the  Prstarea,  it  will  give  the  tabular  area  for  the 
whole  quotient ;  which  must  be  multiplied  by  the  square  of 
the  diameter,  as  before. 


MENSURATION    OF   SUPERFICIES.  105 


EXAMPLES. 

1.  What  is  the  area  of  the  segment  ^ 
of  a  circle  whose  versed  sine,  C  D,  is  ^  / 
2  yards,  and  the  .diameter  of  the  circle,      / 
C  E,  52  yards  ?  ■ 

Here  (2  -i-  52)  =  .038^^,  the  tabular     \ 
versed  sine.  \ 

And  .009763  the  tab.  segment  to  .038. 

.010148  the  next  do.  .039.  ^   • 

.000385  difference. 

Then  .000385  X  t^  =  .000178 

JO 

.009763  tab.  segment  to  .038. 
.009941  tab.  segment  to  .038,^3. 
Now  (.009941  X  52^)  =  (.009941  x  2704)  =  26.880464 
yards,  the  area  of  the  segment. 

2.  What  is  the  area  of  the  segment  of  a  circle  whose 
height  is  10  feet,  and  the  diameter  of  the  circle  50  feet? 

Here  (10  -i-  50)  =  .2  the  tabular  versed  sine. 
And  .111823,  the  tabular  segment  to  .2. 
Now    (.111823  X  50«)  =  (.111823  x  2500)  =  279.5575 
feet,  the  area  required. 

3.  What  is  the  area  of  the  segment  of  a  circle  whose 
versed  sine  is  3  feet,  and  the  diameter  of  the  circle  8  feet  ? 

Ans.  17.2168  ft. 

4.  What  is  the  area  of  the  segment  of  a  circle  whose 
height  is  15  yards,  and  the  diameter  of  the  circle  75  yards  ? 

Ans.  629.004J3  yds. 

5.  What  is  the  area  of  the  segment  of  a  circle  whose 
height  is  6  feet,  and  the  diameter  of  the  circle  21  feet  ? 

Ans.  81.6582  ft. 

6.  What  is  the  area  of  the  segment  of  a  circle  whose 
versed  sine  is  7  feet,  and  the  diameter  of  the  circle  38  feet  * 

Ans.  143.5104  ft. 


PROBLEM  Xn. 

To  find  the  area  of  a  circular  zone,  or  the  space  included 
between  two  parallel  chords  and  their  intercepted  arcs. 


106 


MENSUKATION    OF   SUPERFICIES. 


Rule.  Find  the  area  of  that  part  of  the  zone  A  BCD, 
which  forms  a  trapezoid,  by  Prob.  I.  p.  80,  and  the  area  of 
the  small  segment  Bn  C  B  by  rule  5,  in  the  last  Problem. 

Then  add  the  area  of  the  trapezoid  to  twice  the  area  of 
the  segment,  and  it  will  give  the  area  of  the  zone. 

1.  Let  (/  =  the  diameter  E  F  ;  C,  c  ==  the  two  parallel 
chords  A  B,  DC;  v  =  the  versed  sine  m  n,  and  6  =  G  H, 
the  breadth  of  the  zone,  we  shall  have 

And, 

2.  When  the  two  parallel  chords  are  equal  to  each  other, 
the  diameter  cZ  or  E  F  will  be  =  v^  (6*  -f  c*) ;  and  the 
versed  sine  mn  =  I  [d  —  c). 

And  when  one  of  the  parallel  chords  is  the  diameter,  the 

which  expressions  will  be  found  of  considerable  use  in  facili- 
tating the  computation  of  the  segment  B  n  C  B. 


versed  sine  mn  =—  d — ^  ^/ 

At  <i 


EXAMPLES. 

1.  The  greater  chord  A  B  is  8  yards, 
the  less  D  C  6  yards,  and  their  perpen- 
dicular distance  G  H  is  7  yards ;  re- 
quired the  area  of  the  zone. 

HereAB  +  DC  =  8  +  6  =  14,and 
G  H  =  7. 

Whence  (14  x  7)  -?-  2  =  49,  the  area 
of  the  trapezoid  A  B  C  D 


Also 


^   C  2  ^4x7-^3 


v/(49  +  50  -{-  1)  =  v/100  =  10  the  diameter, 
And     mn=  —  —~^-^[^ — ^- )   +  \ 

^2 


n- 


2      /     '    ^  4x7 
^50  =  5—3.535533  =  1.464467,  the  versed  sine. 


MENSURATION    OF    SUPERFICIES.  107 

Therefore  1.464467  -^  10  =  .1464467,  the  tabular  versed 
sine. 

Answering  to  which,  when  the  work  is  performed  as  in 
the  last  problem,  we  shall  have  .071349. 

Hence   .071349  x  10«  =  .071349  x  100  = 
7.1349,  the  area  of  the  segment  B  n  C  B. 
7.1349  X  2  =  14.2098,  twice  do. 

49.0000,  area  of  the  trapezoid  A  B  C  D. 
63.2698  yards,  area  of  the  zone. 

2.  One  of  the  parallel  chords  of  a  circular  zone  is  48  feet, 
and  the  other  30  feet,  and  its  breadth  13  feet ;  what  is  the 
area  of  the  zone  ?  Jins.  534.19  ft. 

3.  The  greater  chord  of  a  circular  zone  is  16  yards,  and 
the  less  chord  12  yards,  and  their  perpendicular  distance  2 
yards  ;  what  is  the  area  of  the  zone  ?      Ans.  28.379  yds. 

4.  Supposing  the  greater  chord  of  a  circular  zone  to  be 
20  feet,  and  the  less  15  feet,  and  their  distance  ITs  feet; 
what  is  the  area  of  the  zone  1  Ans.  395.4362  ft. 

5.  Required  the  area  of  a  circular  zone,  each  of  whose 
parallel  chords  are  50  feet,  and  their  perpendicular  distance 
30  feet.  Ans.  1668.7093  ft. 

6.  It  is  required  to  find  the  area  of  a  circular  zone,  the 
greater  chord  of  which,  being  equal  to  the  diameter  of  the 
circle,  is  40  feet,  and  the  less  20  feet.       Ans.  592.086  ft. 


PROBLEM  Xni. 

To  find  the  area  of  a  lune,  or  the  space  included  between 
.the  intersecting  arcs  of  the  two  eccentric  circles. 

Rule.  Find  the  areas  of  the  two  segments,  from  which 
the  lune  is  formed  by  Problem  XL,  rule  5,  page  104,  and 
their  difference  will  be  the  area  of  the  lune. 

If  semicircles  be  described  on  the  three  sides  of  a  right- 
angled  triangle  as  diameters,  the  two  lunes  formed  on  the 
base  and  perpendicular,  taken  together,  will  be  equal  to  the 
area  of  the  triangle. 


108 


MENSURATION    OF   SUPERFICIES. 


If  A  B  C,  or  H,  be  a  right-angled 
triangle,  and  semicircles  be  de- 
scribed on  the  three  sides  as  di- 
ameters, then  will  the  triangle  H 
be  equal  to  the  two  lunes  D  and  E 
taken  together. 

For,  if  from  the  greater  semicircle  ABC  there  be  taken  • 
the  two  segments  F  and  G,  there  will  remain  the  triangle 
H  ;  and  if  the  same  segments  be  taken  from  the  other  two 
semicircles,  there  will  remain  the  lunes  D  and  E :  hence, 
since  the  greater  semicircle  is  equal  to  the  sum  of  the  other 
two,  the  triangle  H  must  be  equal  to  the  sum  of  the  lunes 
DandE. 


EXAMPLES. 

1.  The  length  of  the  chord  A  B  is  40 
feet,  the  height  D  C  10  feet,  and  D  E 
4  feet ;  required  the  area  of  the  lune. 

Here  (A  D«  +  C  D'')  -^  C  D  = 
(20"  -f  10*)  -T- 10  =  (400  -f  100)  ■- 
10  =  500  -r-  10  =  50,  the  diameter  be- 
longing to  the  circle  of  which  A  C  B  is  a  part. 

Again  (A  D«  +  D  E^)  -f-  DE  =  (20='  +  4")  -i-  4  =  (400 
4-  16)  -T-  4  =  416  -T-  4  =  104,  the  diameter  belonging  to  the 
circle  of  which  A  E  B  is  a  part. 

Whence  10  -i-  50  =  .2  the  first  tabular  versed  sine. 

Answering  to  which  is  .111823,  the  first  tabular  segment. 

Therefore  .111823  x  50^=.  11 1823  x  2500  =  279.5575, 
the  area  of  the  segment  A  B  C  A. 

Also  4  -i-  104  =.038-j?3-,  the  second  tabular  versed  sine. 

Answering  to  which  is  .009941,  the  second  tabular  seg- 
ment. 

Consequently,  .009941  x  104^  =  .000941  x  10816  = 
107.521856,  the  area  of  the  segment  A  E  B  A. 

Whence  279.5575  —  107.521856  =  172.035644  feet, 
area  of  tne  ;une  A  C  B  E  A. 

2.  The  chord  is  48  feet,  and  the  height  of  the  segments 
18  and  12  feet ;  required  the  area  of  the  lune. 

^ns.  233.8122  ft. 


MENSURATION    OF    SUPERFICIES.  109 

3.  The  chord  is  40  feet,  and  the  height  of  the  segments 
20  and  4  feet ;  required  the  area  of  the  lune. 

^ns.  520.7965  ft. 

4.  Supposing  the  length  of  the  chord  to  be  96  feet,  and 
the  height  of  the  segments  to  be  36  and  14  feet ;  what  is  the 
area  of  the  lune  ?  ^ns.  1634.4350  ft. 

5.  The  length  of  the  chord  is  50  feet,  and  the  heights  of 
^he  segments  18  and  15  feet ;  required  the  area  of  the  lune. 

^ns.  123.8785  ft. 

6.  The  length  of  the  chord  is  40  feet,  and  the  heights  of 
.he  segments  10  and  8  feet ;  what  is  the  area  of  the  lune? 

^ns.  59.5485  ft. 


PROBLEM  XIV. 

To  find  the  area  of  a  part  of  a  ring,  or  of  the  segment  of  a 
sector. 

Rule. — Multiply  half  the  sum  of  the  bounding  arcs  by 
their  distance  asunder,  and  the  product  will  give  the  area. 

EXAMPLES. 

1.  Let  A  B  be  50  inches,  and  D  C 
30  inches,  and  the  distance  D  A  10 
mches  ;  what  is  the  area  of  the  space 
ABC  D? 

Here  [(50  +  30)  -h  2]  X  10  =  (80 
-r-  2)  X  10  =  40  X  10  =  400  inches. 

2.  Let  A  B  =  60  inches,  D  C  = 
iO  inches,  and  D  A  =  2 ;  required 
Ihe  area  of  the  space  A  B  C  D.  ^ns.  100  in. 

3.  Let  A  B  =  25  feet,  D  C  =  1 5  feet,  and  D  A  =  6  feet 
required  the  area  of  the  segment  of  the  sector. 

^ns.  120  ft. 


10 


CONIC   SECTIONS. 


DEFINITIONS. 

« 

§  15.  1-  The  conic  sections  are  certain  plane  figures 
foi  ned  by  the  cutting  of  a  cone,  which  are  of  great  use  in 
some  of  the  higher  branches  of  mathematics. 

2.  A  cone  is  a  solid  described  by  the 
revolution  of  a  right-angled  triangle  about 
one  of  its  legs,  which  remains  fixed ;  as 
ABC. 

3.  The  axis  of  the  cone  is  the  right  line 
about  which  the  triangle  revolves.  c 

4.  The  base  of  a  cone  is  the  circle  which  is  described  by 
the  revolving  leg  of  the  triangle ;  and  its  eJtitude  is  a  per^ 
pendicular  drawn  from  the  vertex  to  the  base ;  as  A  n. 


5.  If  a  cone  be  cut  through  the  vertex  by 
a  plane,  perpendicular  or  oblique  to  that  of 
the  base,  the  section  will  be  a  triangle  ;  as 
ABC. 


ti.  If  a  cone  be  cut  into  two  parts,  by  a 
plane  parallel  to  the  base,  the  section  will 
be  a  circle,  as  A  C  B  D. 


7.  If  a  cone  be  cut  by  a  plane  which 
passes  obliquely  through  its  two  slant  sides, 
the  section  will  be  an  ellipse,  as  A  C  B  D. 

110 


/J^l^ 


CONIC   SECTIONS. 


Ill 


8.  If  a  cone  be  cut  ty  a  plane,  which  is 
parallel  to  either  of  its  slant  sides,  the  sec- 
tion will  be  a  parabola ;  as  A  B  C  A. 


/1^ 


9.  If  a  cone  be  cut  into  two  parts,  by  a 
plane,  which,  being  continued,  would  meet 
the  opposite  cone,  the  section  is  called  an 
hyperbola ;  as  A  B  C. 

The  two  opposite  cones,  in  this  definition, 
are  supposed  to  be  generated  together,  by 
the  revolution  of  the  same  line. 

It  may  here  also  be  observed,  that  all 
figures  which  can  possibly  be  formed  by  the 
cutting  of  a  cone,  are  mentioned  in  these  definitions,  and  are 
the  five  following :  viz.,  a  triangle,  a  circle,  an  ellipsis,  a 
parabola,  and  an  hyperbola;  but  the  last  three  only  are 
usually  called  the  conic  sections. 

10.  If  two  lines  be  drawn  through  the 
centre  of  an  ellipse,  perpendicular  to 
each  other,  and  terminated  both  ways  by 
the  circumference,  the  longest  of  them, 
A  B,  is  called  the  transverse  diameter, 
or  axis,  and  the  shortest,  C  D,  the  con- 
jugate. 

11.  An  ordinate   of  an   ellipse  is  a 
right   line  E  F,  or  E  G,  drawn  from 
any  point  E,  in  the  curve,  perpendicu-  k\ 
lar  to  either  of  the  diameters. 

12.  An  abscissa  is  that  part,  A  F,  or 
C  G,  of  the  diameter  which  is  contain- 
ed between  either  of  the  extremities  of  that  diameter  and  the 
ordinate. 

An  abscissa  may  be  more  generally  considered  as  any 
part  of  the  diameter  or  axis  of  a  curve  comprised  between 
any  fixed  point,  from  which  all  the  abscissas  are  supposed  to 
take  their  origin,  and  another  line  called  the  ordinate,  drawn 
so  as  to  make  a  given  angle  with  the  former,  and  terminated 
in  the  curve ;  but  when  not  otherwise  specified,  they  aro 
commonly  taken  as  above.     The  abscissa  and  its  correspond- 


112  CONIC   SECTIONS. 

iiiff  ordinates,  when  considered  together,  are  also  frequentlj 
called  co-ordinates. 

13.  The  axis  of  a  parabola,  BAG, 
is  a  right  line,  A  D,  drawn  from  the 
vertex,  so  as  to  divide  the  figure  into  two 
equal  parts. 

14.  An  ordinate  of  a  parabola  is  a  right 
line  E  F,  drawn  from  any  point  in  the 
curve,  perpendicular  to  the  axis. 

15.  An  abscissa  is  that  part  of  the  axis,  A  F,  which  is  con- 
tained between  the  vertex  of  the  curve  and  the  ordinate. 

16.  The  transverse  diameter  of  an  hyperbola  is  that  part 
of  the  axis  which  is  intercepted  between  the  two  opposite 
cones,  as  a  B,  in  the  figure  accompanying  definition  ninth. 

17.  The  conjugate  diameter  is  a  line  drawn  through  the 
centre,  perpendicular  to  the  transverse. 

18.  An  ordinate  of  an  hyperbola  is  a  line  drawn  from  any 
point  in  the  curve,  perpendicular  to  either  of  the  diameters ; 
and  an  abscissa  is  that  part  of  the  diameter  which  is  con- 
tained between  either  of  the  extremities  of  that  diameter  and 
the  ordinate. 

Hence,  in  the  ellipse  and  hyperbola,  every  ordinate  has 
two  abscissas,  but  in  the  parabola  only  one,  the  other  vertex 
of  the  diameter  being  at  an  infinite  distance. 

19.  The  parameter  of  any  diameter  is  a  third  proportional 
to  that  diameter  and  its  conjugate. 

20.  The  focus  of  a  conic  section  is  the  point  in  the  axis 
where  the  ordinate  is  equal  to  half  the  parameter,  as  F,  in 
the  figure  accompanying  definition  eleventh,  where  E  F  is 
equal  to  the  semi-parameter  of  the  section. 

The  ellipse  and  hyperbola  have  each  two  foci,  as  F,/:  in 
the  figure  accompanying  the  following  problem,  the  parabola 
has  but  one. 

21.  The  vertices  of  a  conic  section  are  the  points  wliere 
the  cutting  plane  meets  the  opposite  sides  of  the  cone,  or  the 
sides  of  the  vertical  triangular  section.  Hence  the  ellipse 
and  the  opposite  hyperbolas  have  each  two  vertices,  as  A,  B, 
and  B,c,  in  the  figures  accompanying  definitions  seventh  and 
ninth  ;  but  the  parabola  only  one,  as  A,  in  the  figure  accom- 
panying definition  eighth. 


CONIC    SECTIONS. 


113 


THE  ellipse: 


PROBLEM  I. 


§  16.  To  describe  an  ellipse,  the  transverse  and  conjugate 
diameters  being  given. 

Construction.  1.  Draw  the  trans- 
verse and  conjugate  diameters,  A  B, 
C  D,  bisecting  each  other  perpendicu- 
larly in  the  centre  o. 

2.  With  the  radius  A  o,  and  centre 
C,  describe  an  arc  cutting  A  B  in  F,/; 
and  these  tw^o  points  will  be  the  foci  of  the  ellipse. 

3.  Take  any  number  of  points  n,  n,  &c.,  in  the  transverse 
diameter  AB,  and  with  the  radii  A  n,  n  B,  and  centres  F,/, 
describe  arcs  intersecting  each  other  in  «,  s,  &c. 

4.  Through  the  points  s,  s,  &c.,  draw  the  curve  A  sC  sBD, 
and  it  will  be  the  circumference  of  the  ellipse  required. 

It  is  a  well-known  property  of  the  eUipse,  that  the  sum 
of  two  lines  drawn  from  the  foci,  to  meet  in  any  point  in 
the  curve,  is  equal  to  the  transverse  diameter ;  and  from  this 
the  truth  of  the  construction  is  evident. 

From    the    same    principle    is 
derived  another  method  of  describ- 
ing an  eUipse,  by  means  of  a  string  a[ 
and  two  pins. 

Having  found  the  foci  F,/,  as 
before,  take  a  thread  of  the  length 
of   the   transverse    diameter,   and 

fasten  its  ends  with  two  pins  in  the  points  F,/;  then  stretch 
the  thread  F  sf  to  its  greatest  extent,  and  it  will  reach  to  the 
point  8  in  the  curve  ;  and  by  moving  a  pencil  round  within 
the  thread,  keeping  it  always  stretched,  it  will  trace  out  the 
curve  required. 

10* 


114  CONIC   SECTIONS. 


PROBLEM  II. 


In  an  ellipse,  any  three  of  the  four  following  terms  being 
given,  viz.,  the  transverse  and  conjugate  diameters,  an  ordi- 
nate and  its  abscissa,  to  find  the  fourth. 

Case  1. — When  the  transverse,  conjugate,  and  abscissa 
are  given,  to  find  the  ordinate. 

Rule. — As  the  transverse  diameter  is  to  the  conjugate,  so 
is  the  square  root  of  the  product  of  the  two  abscissas,  to  the 
ordinate  which  divides  them. 

It  may  be  remarked  that  if  but  one  abscissa  is  given,  by 
subtracting  it  from  the  transverse  diameter,  we  obtain  the 
other. 

EXAMPLES. 

1 .  In  the  ellipse  A  D  B  C,  the  trans- 
verse diameter  A  B  is  50,  the  conjugate 
diameter  C  D  is  30,  and  the  abscissa 
B  E  18  ;  what  is  the  length  of  the  or- 
dinate E  F  ? 

Here  A  B  =  50,  C  D  =  30,  B  E  = 

18,  and  A  E  =  50  —  18  =  32. 

Whence  50:  30 : :   ^/  (18  x  32)  :  E  F.      Or,  E  F  = 

30  3  72 

—  */576  =  -  X  24  =  — -  =  14.4,  the  ordinate  required. 

50  5  5 

2.  If  the  transverse  diameter  be  40,  the  conjugate  30,  and 
the  abscissa  24,  what  is  the  ordinate  ?  Ans.  14.6969. 

3.  If  the  transverse  diameter  be  120,  the  conjugate  40, 
and  the  abscissa  24,  what  is  the  ordinate  ?  Ans.  16. 

4.  If  the  transverse  diameter  be  35,  the  conjugate  25,  and 
he  abscissa  28,  what  is  the  ordinate  ?  Ans.  10. 

Case  2. — When  the  transverse,  conjugate,  and  ordinate 
are  given,  to  find  the  abscissa. 

Rule.-'— As  the  conjugate  diameter  is  to  the  transverse,  so 
is  the  square  root  of  the  difference  of  the  squares  of  the  ordi- 
nate and  semi-conjugate  to  the  distance  between  the  ordinate 
and  centre. 


CONIC    SECTIONS.  Il5 

And  this  distance  being  added  to,  and  subtracted  from,  the 
semi-transverse,  will  give  the  two  abscissas  required. 

EXAMPLES. 

1.  The  transverse  diameter  A  B  is  60,  the  conjugate 
diameter  C  D  40,  and  the  ordinate  F  E  12;  what  is  the 
length  of  each  of  the  two  abscissas  B  E  and  A  E  ? 

Here  A  B  =  60,  C  D  =  40,  and  F  E  =  12. 
Whence  40 :  60 : :  ^/(20*—  12=) :  O  E. 

Or  O  E  =  ^  v/(400~144)=|^256  =  |  x  16  =  ^ 

=  24. 

And  80  —  24  =  6  =  B  E,  and  30  +  24  =  54  =A  E. 

2.  What  are  the  two  abscissas  to  the  ordinate  10,  the 
diameters  being  35  and  25  ?  Jlns.  7  and  28. 

3.  What  are  the  two  abscissas  to  the  ordinate  16,  the 
diameters  being  120  and  40  ?  Jlns.  24  and  96. 

4.  What  are  the  two  abscissas  to  the  ordinate  14.4,  the 
diameters  being  50  and  30  ?  jSns.  18  and  32. 

Case  3. —  When  the  conjugate,  ordinate,  and  abscJsssa  are 
given,  to  find  the  transverse  diameter.  ^ 

Rule. — To,  or  from  the  semi-conjugate,  according  as  the 
less  or  greater  abscissa  is  used,  add  or  subtract  the  squaio 
root  of  the  difference  of  the  squares  of  the  ordinate  and  semi- 
conjugate. 

Then,  as  the  square  of  the  ordinate  is  to  the  product  of  the 
conjugate  and  abscissa,  so  is  the  sum  or  difference  above 
found  to  the  transverse  diameter  required. 

EXAMPLES. 

1.  The  conjugate  diameter  C  D  is  60,  the  ordinate  E  F 
24,  and  the  less  abscissa  B  E  36  ;  required  the  transverse 
diameter  A  B. 

Hete  O  C  =  30,  E  F  =  24,  and  B  E  =  36. 

Whence  30  +  ^  (30^—24^)  =  30  +v/(900— 576)=^0 
+  ^  324  =  30  +  18  =  48. 

And  24^:  (00  x  36)  : :  48:  (60  x  36  x  48)  -r-  24'  = 
103()80  -i-  576  =  ISO,  the  transverse  diameter  reauired. 


116  CONIC    SECTIONS. 

2.  The  conjugate  diameter  is  60,  the  ordinate  18,  and  the 
less  abscissa  24 ;  required  the  transverse  diameter. 

Ans.  240. 

3.  The  conjugate  diameter  is  40,  the  ordinate  16,  and  the 
fnreater  abscissa  96 ;  required  the  transverse  diameter. 

Ans.  120. 

4.  The  conjugate  diameter  is  2.5,  the  ordinate  10,  and  the 
greater  abscissa  28 ;  required  the  transverse  diameter. 

Ana.  35. 

Case  4.  The  transverse,  ordinate,  and  abscissa  being 
given,  to  find  the  conjugate  diameter. 

Rule. — As  the  square  root  of  the  product  of  the  two 
abscissas  is  to  the  ordinate,  so  is  the  transverse  diameter  to 
the  conjugate. 

EXAMPLES. 

1 .  The  transverse  A  B  is  82,  the  ordinate  E  F  8,  and  the 
abscissa  B  E  32 ;  required  the  conjugate  diameter  C  D. 

Here  A B  =  82,  EF  =  8,  BE  =  32,  and  AE  =  82 
—  32  =  50. 

Hence  ^  (32  x  50) :  8  : :  82  :  (8  x  82)  -4-  x/  (32  x  50) 
=  656  -j-  x/1600  =  656  -=-  40  =  16.4,  the  conjugate  diame- 
ter required. 

2.  The  transverse  diameter  is  35,  the  ordinate  10,  and  the 
abscissa  28 ;  what  is  the  conjugate  ?  Ans.  25. 

3.  The  transverse  diameter  is  120,  the  ordinate  16,  and 
its  abscissa  24 ;  what  is  the  conjugate  ?  Ans.  40. 

4.  The  transverse  diameter  is  50,  the  ordinate  14.4,  and 
the  abscissa  32 ;  what  is  the  conjugate  ?  Ans.  30. 


PROBLEM  m. 

The  transverse  and  conjugate  diameters  being  given,  to 
find  the  circumference. 

Rule. — Multiply  the  square  root  of  half  the  sum  of  the 
squares  of  the  two  diameters  by  3.1416,  and  the  product 
«vill  be  the  circumference  nearly. 


CONIC   SECTIONS. 


m 


EXAMPLES. 

1.  The  transverse  is  40,  and  the 
conjugate  80 ; '  required  the  circum- 
ference of  the  ellipse. 

Here  ^  [(40^  +  30»)  -r-  2]  = 
/  >/  [(1600  +  900)  -T-  2 ]  = 

^/  (2500  ->  2)  =  v/  1250 
=  35.3553. 

Whence  35.3553  x  3.1416  =  111.0722  =  the  circura 
ference. 

2.  The  transverse  diameter  is  24,  and  the  conjugate  20; 
required  the  circumference  of  the  ellipse.    Ans.  69.4001. 

3.  The  transverse  diameter  is  60,  and  the  conjugate  40 ; 
what  is  the  circumference?  ^ns.  160.1907. 

4.  The  transverse  diameter  is  24,  and  the  conjugate  18 ; 
what  is  the  circumference  ?  Jins.  66.6433. 


PROBLEM  rV. 

The  transverse  and  conjugate  diameters  being  given,  to 
find  the  area. 

Rule. — Multiply  the  transverse  diameter  by  the  conjugiiie, 
and  this  product  again  by  .7854,  and  the  resxilt  will  be  the 
area. 

EXAMPLES. 

1.  Required  the  area  of  an  ellipse  whose  transverse  di- 
ameter is  24  perches,  and  the  conjugate  18  perches. 

Here  24  x  18  x  .78M  =  339.2928  perches  =  2  a.  0  r. 
19.2928  p.,  the  area  required. 

2.  Required  the  area  of  an  ellipse  whose  two  diameters 
are  60  and  40  rods.  ^ns.  11  a.  3  r.  4.96  p. 

3.  Required  the  area  of  an  eUipse  whose  two  diameters 
are  40  and  36  chains.  ^ns.  113  a.  0  r.  15.616  p. 

4.  The  transverse  and  conjugate  diameters  are  66  and  22 
yards ;  what  is  the  area?  J3ns.  1140.4  yds. 


US  coiric  SECTioirs. 


PROBLEM  V. 

The  transverse  and  conjugate  diameters  of  an  ellipse  being 
given,  to  find  the  diameter  of  a  circle  containing  the  same 
area. 

Rtjle. — ^Multiply  the  transverse  and  conjugate  diameters 
together,  and  extract  the  square  root  of  their  product. 


EXAMPLES. 

1.  The  transverse  and  conjugate  diameters  are  70  and  50 
chains ;  what  is  the  diameter  of  a  circle  containing  the  same 
area? 

Here  ^  (70  x  50)  =  v/  3500  =  59.16079  chains,  the 
diameter. 

2.  The  transverse  and  conjugate  diameters  are  36  and  24 
chains;  what  is  the  diameter  of  a  circle  containing  the  same 
area  ?  Jns.  29.3938  chs. 

3.  The  transverse  and  conjugate  diameters  are  49  and  16 
rods ;  what  is  the  diameter  of  a  circle  containing  the  same 
area  ?  J3ns.  28  rods. 

4.  The  transverse  and  conjugate  diameters  are  87  and  52 
feet ;  what  is  the  diameter  of  a  circle  containing  the  same 
area  ?  ^ns.  67.2606  ft. 


PROBLEM  VL 

To  find  the  area  of  an  elliptic  segment  whose  base  is 
parallel  to  either  of  the  axes  of  the  ellipse. 

Rule. — Divide  the  height  of  the  segment  by  that  axis  of 
the  ellipse  of  which  it  is  a  part,  and  find  in  the  table,  p.  295, 
a  circular  segment  whose  versed  sine  is  equal  to  the  quotient. 

Then,  multiply  the  segment  thus  found  and  the  two  axes 
of  the  elUpse  together,  and  the  product  will  be  the  area 
required. 


CONIC    SECTION  a.  119 


EXAMPLES. 

1.  Required  the  area  of  the  elliptic 
segment  whose  height  A  G  is  20  chains, 
and  the  axes  of  the  ellipse,  2  A  B  and 
C  D,  70  and  50  chains  respectively. 

Here  20  -r-  70  =  .2854,  the  tabular 
versed  sine.     And  the  tabular  segment     ;  i  ; 

belonging  to  this,  as  found  byProb.XI.,  c' -j^ -d 

Rules,  p.  104,  is  .1851(56. 

Whence  .185166  x  70  (2AB)  x  50  (CD)  =  648.081 
chains  =  64  a.  3  r.  9.296  p.,  the  area  of  the  segment  E  A  F. 

2.  What  is  the  area  of  an  elliptic  segment  cut  off  by  an 
ordinate  parallel  to  the  transverse  diameter  whose  height  is 
20  feet,  the  axes  being  80  and  50  feet  ? 

Ms.  1173.476  ft. 

3.  What  is  the  area  of  an  elliptic  segment  cut  off  by  an 
ordinate  parallel  to  the  transverse  diameter  whose  height  is 
10  chains,  the  axes  being  35  and  25  chains  ? 

Ms.  25  a.  2  r.  27.166  p. 

4.  What  is  the  area  of  an  elliptical  segment  cut  off  by  an 
ordinate  parallel  to  the  conjugate  diameter  whose  height  is 
10  feet,  the  axes  of  the  ellipse  being  35  and  25  feet  ? 

^ns.  102.0202  ft. 

5.  What  is  the  area  of  an  elliptic  segment  cut  off  by  an 
ordinate  parallel  to  the  transverse  diameter  whose  height  is 
5  yards,  the  axes  being  35  and  25  yards  ? 

^ns.  97.8451  yds. 

6.  What  is  the  area  of  an  elliptic  segment  cut  off  by  an 
ordinate  parallel  to  the  conjugate  diameter,  the  axes  of  the 
ellipse  being  60  and  40  feet,  and  the  height  of  the  segment 
15  feet?  ^n*.  368.51  ft, 

7.  What  is  the  area  of  an  elliptic  segment  cut  off  by  a 
double  ordinate  parallel  to  the  conjugate  axis,  at  the  distance 
of  36  yards  from  the  centre,  the  axes  being  120  and  40  yards  ? 

^ns.  536.7504  yds. 


120 


CONIC    SECTIONS. 


THE  PARABOLA. 


PROBLEM  I. 


§  17.  To  describe  a  parabola,  any  ordinate  to  the  axis 
and  its  abscissa  being  given. 

Construction.  1 .  Let  V  R  and  R  S  be 
the  given  abscissa  and  ordinate ;  bi- 
sect the  latter  in  m,  join  Vm,  and  draw 
m  n  perpendicular  to  it,  meeting  the  axis 
in  n. 

2.  Make  V  C  and  V  F  each  equal  to 
Rn,  and  F  will  be  the  focus  of  the  curve. 

3.  Take  any  number  of  points  r,  r,  &c. 
in  the  axis,  through  which  draw  the 
double  ordinates-S  r  S,  &c.,  of  an  indefi- 
nite length. 

4.  With  the  radii  C  F,  Cr,  &c,,  and  centre  F,  describe 
arcs  cutting  the  corresponding  ordinates  in  the  points  s,  s, 
&c.,  and  the  curve  S  V  S  drawn  through  all  the  points  of  in- 
tersection will  be  the  parabola  required. 

The  Une  sF s  passing  through  the  focus  F  is  called  the 
parameter. 


C 

Yi 

Kfi 

t 

/ 

'"r 

r 

r 

\    A 

R 

\     \ 

I.--' 
n 


PROBLEM  n. 

In  a  parabola,  any  three  of  the  four  following  terms  being 
given,  viz.,  any  two  ordinates  and  their  two  abscissas,  to  find 
the  fourth. 


Rule. — ^As  any  abscissa  is  to  the  square  of  its  ordinate,  so 
is  any  other  abscissa  to  the  square  of  its  ordinate.  Or  as  the 
square  root  of  any  abscissa  is  to  its  ordinate,  so  is  the  square 
root  of  any  other  abscissa  to  its  ordinate,  and  conversely. 


CONIC    SECTIONS. 


121 


EXAMPLES. 

The  abscissa  V  F  is  18,  and  its  or- 
dinate E  F  13 ;  required  the  ordinate 
G  H,  the  abscissa  of  which,  V  H,  is  32. 

Here  VF  =  18,  EF  =  12,  and 
VH=32. 

12*  X  32 


Whence  18:  12'::  32 
144x32 


18 


18 


=  256  =:  the  square  of  the 


ordinate  G  H,  and  y/  256  =  16  =i  the  ordinate  required. 

2.  The  abscissa  VF  is  25  and  its  ordinate  EF  16; 
required  the  ordinate  G  H,  the  abscissa  of  which,  V  H. 
is  49. 

U  yoR     1A        ,Ac^    16v/49        16x7       112 

Here   ^25  :  16::  ^  49:  — —  = -^- =.    -  = 

22.4  =  the  ordinate  G  H. 

3.  The  abscissa  V  F  is  9,  and  the  ordinates  E  F  6  and 
GH  8 ;  required  the  abscissa  VH. 

8^  X  9       64  X  9 


Here  6^ :  8^ : :  9 
abscissa  V  H. 


6^ 


36 


64 -T- 4  =  16  =  the 


4.  The  abscissa  VH  is  49,  and  its  ordinate  GH  22.4 ;  re- 
quired the  ordinate  E  F,  the  abscissa  of  which,  V  F,  is  25. 

Ans.  16,  the  ordinate  E  F. 


PROBLEM  m. 


To  find  the  length  of  any  arc  of  a  parabola  cut  off  by 
double  ordinate. 

Rule. — To  the  square  of  the  ordinate  add  4  of  the  square 
of  the  abscissa,  and  twice  the  square  root  of  the  sum  will  be 
the  length  of  the  arc  nearly. 


11 


122 


CONIC   SECTIONS. 


EXAMPLES. 

1.  The  abscissa  VH  is  3,  and  its 
ordinate  G  H  6  ;  what  is  the  length  of 
the  arc  GVK? 

Here  6^+|  X  3'  =36  +1  x  9  = 

36  +  12  =  48.  And  ^  (48)  x  2  = 
6.9282  X  2  =  13.8564  =  the  length 
of  the  arc  GVK. 

2.  The  abscissa  VH  is  6,  and  its  ordinate  GH  12;  re- 
quired the  length  of  the  arc  GVK.  ^ns.  27.7128. 

3.  The  abscissa  is  15,  and  its  ordinate  8 ;  what  is  the 
length  of  the  arc  ?  Ans.  38.1575. 

4.  The  abscissa  is  9,  and  its  ordinate  6 ;  what  is  the 
length  of  the  arc  ?  Ans.  24.    • 


PROBLEM  IV. 

To  find  the  area  of  a  parabola,  its  base  and  height  being 
given. 

Rule. — Multiply  the  base  by  the  height,  and  two-thirds 
of  the  product  will  be  the  area  required. 


EXAMPLES. 

1.  What  is  the  area  of  the  parabola 
BAG,  whose  height  A  D  is  12,  and 
the  base,  or  double  ordinate,  BC,  30? 

Here  B  C  =  30,  and  A  D  =  12. 

Whence  (30  x  12)  x  ^  =  360  x  | 
o  o 

=  240  =  area  required. 

2.  The  abscissa  is  18,  and  the  base,  or  double  ordinate, 
42 ;  what  is  the  area  ?  Ans.  504. 

3.  What  is  the  area  of  a  parabola  whose  abscissa  is  1 1, 
and  its  double  ordinate  21  ?  Ans.  154. 

4.  What  is  the  area  of  a  parabola  whose  height  is  9,  and 
the  base,  or  double  ordinate,  24  ?  Ans.  144. 


CONIC    SECTIONS. 


123 


PROBLEM  V. 

To  find  the  area  of  a  frustum  of  a  parabola. 

Rule. — Divide  the  difference  of  the  cubes  of  the  two  ends 
of  the  frustum  by  the  difference  of  their  squares,  and  this 
quotient,  multipHed  by  two-thirds  of  the  altitude,  will  give 
the  area  required. 

EXAMPLES. 

A 

1.  In  the  parabolic frustumBEGD, 
the  two  parallel  ends,  B  D  and  E  G, 
are  6  and  10,  and  the  altitude,  or  part 
of  the  abscissa,  C  F,  is  3  ;  what  is  the 
area? 

Here  B  D  =  6,  E  G  =  10,and  C  F 
=  3. 

Whence  (10'  —  6»)  ^  (10'  —  6^)  = 
(100*0  —  2HJ)~  (100  —  36)  =  784  -=-  64  =  12.25. 

Then  12.25  X  {tz  of  3)  =  12.25  x  2  =  24.50,  the  area. 
o 

2.  Required  the  area  of  a  parabolic  frustum,  the  greater 
end  of  which  is  10,  the  less  6,  and  the  height  4.2. 

^ns.  34.3. 

3.  The  greater  end  of  the  frustum  is  12,  the  less  10,  and 
the  height  2.4 ;  what  is  the  area  ?  ^ns.  26f  f . 

4.  The  greater  end  of  the  frustum  is  24,  and  the  less  20, 
and  the  altitude  6  ;  what  is  the  area  ?        ^ns.  132.3636. 

5.  Required  the  area  of  the  parabolic  frustum,  the  greater 
end  of  which  is  10,  the  less  6,  and  the  height  5. 

^ns.  40.81. 

6.  The  greater  end  of  the  frustum  is  24,  the  less  end  20, 
and  the  height  5i  ;  what  is  the  area  ?  ^ns.  121 3. 

7.  Required  the  area  of  the  parabolic  frustum,  the  greater 
end  of  which  is  10,  the  less  6,  and  the  height  4. 

.5n«.  32|. 


124  .  CONIC    SECTIONS. 

THE  HYPERBOLA. 
PROBLEM  I. 

§  1 8.  To  construct  an  hyperbola,  the  transverse  and  con- 
jugate diameters  being  given. 

Construction.  1.  Make  A  B  the 
transverse  diameter,  and  C  D  perpen- 
dicular to  it,  the  conjugate. 

2.  Bisect  A  B  in  O,  and  from  O,  with     /  ^ 
the  radius  O  C  or  O  D,  describe  the  cir-  •'  \  a    o* 
cle  D  /  C  F,  cutting  A  B  produced  in 
F  and/,  which  points  will  be  the  two 
foci. 

3.  In  A  B  produced,  take  any  num- 
ber of  points,  n,  n,  &c.,  and  from  F  and/,  as  centres,  with 
the  distances  B  n,  A  n,  as  radii,  describe  arcs  cutting  each 
other  in  s,  s,  &c. 

4.  Through  the  several  points  s,  s,  &c.,  draw  the  curve 
»  B  *,  and  it  will  be  the  hyperbola  required. 

If  straight  lines  be  drawn  from  the  point  O,  through  the 
extremities  C,  D  of  the  conjugate  diameter  C  D,  they  will  be 
the  asymptotes  of  the  hyperbola,  whose  property  it  is  to  ap- 
proach continually  to  the  curve,  without  ever  meeting  it. 


PROBLEM  II. 

In  an  hyperbola,  any  three  of  the  four  following  terms 
being  given,  viz.,  the  transverse  and  conjugate  diameters,  an 
ordinate,  and  its  abscissa,  to  find  the  fourth. 

Case  1.  The  transverse  and  conjugate  diameters,  and  the 
two  abscissas  being  given,  to  find  the  ordinate. 

Rule. — As  the  transverse  diameter  is  to  the  conjugate,  so 
is  the  square  root  of  the  product  of  the  two  abscissas  to  the 
ordinate  required. 

It  may  be  observed  that  in  the  hyperbola  the  less  abscissa 
added  to  the  transverse  diameter  gives  the  greater,  and  the 
transverse  diameter  subtracted  from  the  greater  abscissa 
gives  the  less. 


CONIC    SECTIONS. 


125 


EXAMPLES. 

1.  In  the  hyperbola  BAG, 
the  transverse  diameter  is  50, 
the  conjugate  80,  and  the  less 
abscissa  AD,  12;  required  the 
ordinate  D  C. 

Here   transverse  diameter  = 
50,  conjugate  =  30,  A  D  =  12,  and  12  +  50  =  62,=  the 
greater  abscissa. 

When  ;e  50 :  30  :  :  v/(62  x  12)  :  ^  v/  (62  x  12)  =  ^ 

3 

^/744  =  ^  X  27.27636  =  16.3658  =  the  ordinate. 
5 

2.  The  transverse  diameter  is  24,  the  conjugate  21,  and 
the  less  abscissa  8;  what  is  the  ordinate  ?  Ans.  14. 

3.  The  transverse  diameter  is  36,  and  the  conjugate  24, 
and  the  less  abscissa  12;  what  is  the  ordinate?     Ans.  16. 

4.  The  transverse  diameter  is  120,  the  conjugate  72,  and 
greater  abscissa  160 ;  what  is  the  ordinate  ?         Arts.  48. 

Case  2. — The  transverse  and  conjugate  diameters,  and  an 
ordinate  being  given,  to  find  the  two  abscissas. 

Rule. — ^As  the  conjugate  diameter  is  to  the  transverse,  so 
is  the  square  root  of  the  sum  of  the  squares  of  the  ordinate 
and  semi-conjugate  to  the  distance  between  the  ordinate  and 
centre,  or  half  the  sum  of  the  abscissas. 

Then  the  sum  of  this  distance  and  the  semi-transverse 
will  give  the  greater  abscissa,  and  their  difference  will  give 
the  less. 

EXAMPLES. 

1.  The  transverse  diameter  is 
36,  the  conjugate  24,  and  the  or- 
dinate, B  U,  16 ;  what  are  the 
two  abscissas  ? 

Here  ^(16^  -f  12^)  =  -v/(256  ^^ 
+  144)  =  ^  400  =  20. 

Then  24  :  36  .  :  20 :  30  =—  sum  of  the  abscissas. 

And  30  -f  18  =  48  =  the  greater  abscissa. 
Also  30 —  18  =  12  =-  the  less  abscissa. 
11* 


126  CONIC    SECTIONS. 

2.  The  transverse  diameter  is  120,  the  conjugate  72,  and 
the  ordinate  48 ;  what  are  the  two  abscissas  ? 

Ans.  IGO  and  40. 

3.  The  transverse  and  conjugate  diameters  are  24  and 
21 ;  required  the  two  abscissas  to  the  ordinate  14  ? 

^ns.  32  and  8. 

4.  The  transverse  being  60,  and  the  conjugate  36 ;  re- 
quired the  two  abscissas  to  the  ordinate  24. 

Ans.  80  and  20. 

Case  3. — ^The  transverse  diameter,  the  two  abscissas,  and 
the  ordinate  being  given,  to  find  the  conjugate. 

Rule. — As  the  square  root  of  the  product  of  the  two 
abscissas  is  to  the  ordinate,  so  is  the  transverse  diameter  to 
the  conjugate. 

EXAMPLES. 

1.  The  transverse  diameter  is  36,  the  ordinate  16,  and  the 
two  abscissas  are  48  and  12  ;  required  the  conjugate. 

Here  v/(48  x  12)  =  v/576  =24. 

Whence  24  :  16  : :  36 :  24  =  the  conjugate  required. 

2.  The  transverse  diameter  is  60,  the  ordinate  24,  and  the 
two  abscissas  are  80  and  20 ;  required  the  conjugate. 

Ans.  36. 

3.  The  transverse  diameter  is  36,  the  ordinate  21,  and  the 
abscissas  12  and  48 ;  required  the  conjugate.    Ans.  31.5. 

4.  The  transverse  diameter  is  24,  the  ordinate  14,  and  the 
abscissas  8  and  32;  required  the  conjugate.  Ans.  21. 

Case  4. — The  conjugate  diameter,  the  ordinate,  and  the 
two  abscissas  being  given,  to  find  the  transverse. 

Rule  1. — Add  the  square  of  the  ordinate  to  the  square  of 
the  semi-conjugate,  and  find  the  square  root  of  their  sum. 

2.  Take  the  sum  or  difference  of  the  semi-conjugate  and 
this  root,  according  as  the  less  or  greater  abscissa  is  used,  and 
then  say :  As  the  square  of  the  ordinate  is  to  the  product  of  the 
abscissa  and  conjugate,  so  is  the  sum  or  difference  above 
found  to  the  transverse  required. 


CONIC   SECTIONS.  127 


EXAMPLES. 


1.  The  conjugate  diameter  is  21,  the  ordinate  14,  and  the 
less  abscissa  8  ;  required  the  transverse. 

Here  x/(14'  +  10.5^)=  v/(  196  -f  1 10.25)  =  ^/306.25  = 
17.5,  and  17.5  +  10.5  =  28. 

Also  21  X  8==  168. 

Whence  196  (14') :  168  ::  28:  24=  the  transverse  re- 
quired. 

2.  The  conjugate  diameter  is  72,  the  ordinate  48,  and  the 
less  abscissa  40  ;  what  is  the  transverse  ?  Ans.  120. 

3.  The  conjugate  diameter  is  36, the  less  abscissa  20,  and 
the  ordinate  24  ;  required  the  transverse.  Ans.  60. 

4.  The  conjugate  diameter  is  31.5,  the  ordinate  21,  and 
the  gpreater  abscissa  48  ;  required  the  transverse. 

Ans.  36. 


PROBLEM  in. 

To  find  the  length  of  any  arc  of  an  hyperbola,  beginning  at 
the  vertex. 

Rule. — To  19  times  the  square  of  the  transverse  add  21 
times  the  square  of  the  conjugate;  also  to  9  times  the  square 
of  the  transverse  add  as  before  21  times  the  square  of  the 
conjugate  ;  and  multiply  each  of  these  sums  by  the  abscissa. 

2.  To  each  of  the  two  products  thus  found  add  15  times 
the  product  of  the  transverse  and  the  square  of  the  conju- 
gate. 

3.  Then  as  the  less  of  these  results  is  to  the  greater,  so  is 
the  ordinate  to  the  length  of  the  arc  nearly. 

EXAMPLES. 

1.  In  the  hyperbola  BAG, 
the  transverse  diameter  is  80,  the 
conjugate  60,  the  ordinate  B  D, 
10,  and  the  abscissa  A  D,  2. 1637; 
required  the  length  of  the  arc 
BAG. 

Here  2.1637  x  [(19  x  80^)  +  (21  x  60^)]  =  2.1637  x 
(121600  -f  75600)  =  2.1637  x  197200  =  426681.64. 


128  CONIC    SECTIONS. 

And  2. 1637  X  [(9  x  80^ +  (21  x  60^)]=2.163Tx  (57600 
+  75600)=  2.1637  x  133200=  288204.84. 

Whence  (15  x  80  x  60')  +  426681.64  =  4320000  + 
126681.64  =  4746681 .64. 

And  (15x80  x60^)+288204.84=4320000 +288204.84 
r=  4608204.84. 

Then  4608204.84  :  4746681.64  :  :  10  :  10.3005  =  the 
length  of  the  arc  A  C.  Therefore,  10.3005  x  2  =  20.601  = 
the  length  of  the  whole  arc  B  A  C. 

2.  The  transverse  diameter  of  an  hyperbola  is  120,  the 
conjugate  72,  the  ordinate  48,  and  the  abscissa  40 ;  required 
the  whole  length  of  the  curve.  Ans.  125.304. 

3.  The  transverse  diameter  is  80,  the  conjugate  60,  and 
the  ordinate  16 ;  required  the  length  of  the  arc. 

Ans.  17.0856. 

4.  Required  the  whole  length  of  the  curve  of  an  hyperbola, 
to  the  ordinate  20;  the  transverse  and  conjugate  axis  being 
80  and  70.  Ans.  42.267. 


PROBLEM  IV. 

To  find  the  area  of  an  hyperbola,  the  transverse,  conjugate, 
and  abscissa  being  given. 

Rule. — 1.  To  the  product  of  the  transverse  and  abscissa, 
add  f  of  the  square  of  the  abscissa,  and  multiply  the  square 
root  of  the  sum  by  21. 

2.  Add  4  times  the  square  root  of  the  product  of  the  trans- 
verse and  abscissa  to  the  product  last  found,  and  divide  the 
sum  by  75. 

3.  Then  if  4  times  the  product  of  the  conjugate  and  ab- 
scissa be  divided  by  the  transverse,  this  last  quotient  multi 
plied  by  the  former  wiU  give  the  area  required  nearly. 


EXAMPLES. 

1.  In  the  hyperbola  BAG, 
the  transverse  axis  is  100,  the 
conjugate  60,  and  the  abscissa,  or 
height  A  D,  is  50 ;  required  the 
urea. 


CONIC    SECTIONS. 


129 


Here  21  ^  [(100  x  50)  +  (^  X  50*)] 

=  21  X  82.37544710  =  1729.8844. 

And  [4  v/  (100  X  50)  +  1729.8844]  -^  75 

=  [(4  X  70.7107)  +  1729.8844]  -r-  75 

=  (282.8428  +  1729.8844)  ■-  75 

=  2012.7272  -T-  75  =  26.836362. 
Whence  [(4  x  50  x  60)  -f- 100]  X  26.836362  = 

(12000  -=-  100)  X  26.836362  =  120  X  26.836362  = 

3220.3634  =  the  area  required. 

2.  Required  the  area  of  the  hyperbola  to  the  abscissa  25, 
the  two  axes  being  50  and  30.  Ans.  805.0908. 


PROBLEM  V. 

To  find  the  area  of  a  space  A  N  O  B,  bounded  on  one 
side  by  the  curve  of  an  hyperbola,  by  means  of  equidistant 
ordinates. 

Let  A  N  be  divided  into  a  given 
number  of  equal  parts,  A  C,  C  E, 
&c.,  and  let  perpendicular  ordi- 
nates A  B,  C  D,  &c.,  be  erected, 
and  terminated  by  any  hyperbolic 
curve,  B  D  F,  &c. ;  and  let  A  = 
AB  +  NO,  B=CD  +  GH-f 
L  M,  &c.,  and  C  =  E  F  -}- 1  K,  &c. ;  then  the  common  dis- 
tance A  C  of  the  ordinates,  being  multiplied  by  the  sum 
arising  from  the  addition  of  A,  4  B,  and  2  C,  and  one-third 
of  the  product  taken  will  be  the  area  very  nearly. 


EXAMPLES. 

1.  Given  the  lengths  of  9  equidistant  ordinates,  14,  15, 
16,  17,  18,  20,  22,  23,  and  25  feet,  and  the  common  distance 
2  feet ;  what  is  the  area  ? 

Here  [A  C  x  (A  +  4  B  +  2  C)]  -=-  3  =  [2  X  (39  -f  300 
-t-  112)]  ^  3  =  (2  X  451)  -f-  3  =  902  -4-  3  =  300§  feet. 

2.  Given  the  lengths  of  3  equidistant  ordinates,  A  B  = 
5  feet,  C  D  =  7,  and  E  F  =  8,  the  length  of  the  base  A  E  = 
10  ;  what  is  the  area  of  the  figure  A  B  F  E  ?     Jim.  68i  ft. 


MENSURATION  OF  SOLIDS. 


DEFINITIONS. 


§  10.  1.  The  measure  of  any  solid  body  is  the  whole 
capacity  or  ODntent  of  that  body,  when  considered  under  the 
triple  dimensions  of  length,  breadth,  and  thickness. 

2.  A  cube  whose  side  is  one  inch,  one  foot,  or  one  yard, 
&c.,  is  called  the  measuring  unit;  and  the  content  or  solidity 
of  any  figure  is  estimated  by  the  number  of  cubes  of  this 
kind  which  are  contained  in  it. 

A 


3.  A  cube  is  a  solid  contained  by  six  equa 
square  sides,  or  faces,  as  A  B  C  D  E  F. 


4.  A  parallelopipedon  is  a  solid  con-  b 
tained  by  six  rectangular  plane  faces, 
every  opposite  two  of  which  are  equal 
and  parallel,  asABCDEF.  c 

5.  A  prism  is  a  solid  whose  ends  are  two 
equal,  parallel,  and  similar  plane  figures,  and  its 
sides  parallelograms  ;  as  A  B  C  D  E  F. 

It  is  called  a  triangular  prism  when  its  ends 
are  triangles ;  a  square  prism  when  its  ends  are 
squares ;  a  pentagonal  prism,  when  its  ends  are 
pentagons,  and  so  on. 

6.  A  cylinder  is  a  solid  described  by  the  revo- 
lution of  a  rectangle  about  one  of  its  sides  as  an 
axis,  which  remains  fixed ;  as  A  B  C  D. 


130 


MENSURATION    OF    SOLIDS. 

7.  A  cone  is  a  solid  described  by  the 
revolution  of  a  right-angled  triangle  about 
one  of  its  legs,  which  remains  fixed  ;  as 
ABC. 


8.  A  pyramid  is  a  solid  whose  sides  are  all 
triangles  meeting  in  a  point  at  the  vertex,  and 
the  base  any  plane  figure ;  as  A  B  C  D  E,  which 
is  a  pentagonal  pyramid. 

When  the  base  is  a  triangle,  it  is  called  a  fri-  j 
angular  pyramid;  when  a  square,  it  is  called 
a  square  or   quadrangular  pyramid ;  when  a 
pentagon,  it  is  called  a.  pentagonal  pyramid,  &c, 

9.  A  sphere  is  a  solid  described  by  the 
revolution  of  a  semicircle  about  its  diameter, 
which  remains  fixed  ;  as  A  B  C  D.  b|1 

10.  The  centre  of  a  sphere  is  a  point 
within  the  figure,  equally  distant  from  every 
point  of  its  convex  surface. 

11.  A  diameter  of  the  sphere  is  a  straight  line  passing 
through  its  centre,  and  terminated  both  ways  by  the  convex 
surface. 

And  if  it  be  the  diameter  about  which  the  generating 
semicircle  revolves,  it  is  called  the  axis  of  the  sphere. 

12.  A  circular  spindle  is  a  solid 
generated  by  the  revolution  of  a  seg- 
ment of  a  circle  about  its  chord,  which 
remains  fixed  ;  as  A  B  D  C. 

Elliptic,    parabolic,  and  hyperbolic 
spindles  are  generated    in    the    same 
manner  as  circular  spindles,  the  double  ordinate  of  the  sec  • 
tion  being  always  considered  as  fixed. 

13.  A  spheroid  or  ellipsoid  is  a  solid 
generated  by  the  revolution  of  a  semi- 
ellipsis  about  one  of  its  axes,  which  re- 
mains fixed  ;  as  A  B  D  C. 

The  spheroid  is  called  prolate,  when 
the  revolution  is  made  about  the  trans- 
verse axis,  and  oblate  when  it  is  made 
about  the  conjugate  axis. 


132  MENSURATION    OF   SOLIDS. 

14.  Parabolic  and  hyperbolic  conoids 
are  solids  formed  by  the  revolution  of  a 
semi-parabola  or  semi-hyperbola  about  its 
transverse  axis,  which  is  considered  as 
quiescent ;  as  A  B  D ;  and  the  same  for 
any  other  soUd  of  this  kind. 

15.  A  segment  of  a  pyramid,  sphere,  or  of  any  other  solid, 
is  a  part  cut  off  from  the  top  of  it  by  a  plane  parallel  to  the 
base  of  the  figure. 

16.  A.  frustum  or  trunk,  is  the  part  that  remains  at  the 
bottom,  after  the  segment  is  cut  off. 

17.  The  zone  of  a  sphere,  is  that  part  which  is  intercepted 
between  two  parallel  planes ;  and  when  those  planes  are 
equally  distant  from  the  centre,  it  is  called  the  middle  zone 
of  the  sphere. 

18.  The  height  of  a  solid  is  a  perpendicular  drawn  from 
its  vertex  to  the  base,  or  to  the  plane  on  which  it  is  supposed 
to  stand. 

19.  A  wedge  is  a  sohd,  having  a  rectangular  base,  and 
two  of  its  opposite  sides  meeting  in  an  edge. 

20.  A  prismoid  is  a  solid,  having  on  its  ends  two  rect 
angles  parallel  to  each  other,  and  its  upright  sides  are  foui 
trapezoids. 

21.  An  ungula,  or  hoof,  is  a  part  cut  off  from  a  sohd  by  a 
plane  obUque  to  the  base. 

The  surfaces  of  all  similar  solids  are  to  each  other  as  the 
squares  of  their  like  dimensions ;  such  as  diameters,  circum 
ferences,  Hke  linear  sides,  &c.,  &c.,  and  their  solidities,  as 
the  cubes  of  those  dimensions. 

The  solidity  of  cylinders,  prisms,  parallelopipedons,  &c.» 
»vhich  have  their  altitudes  equal,  are  to  each  other  as  the 
squares  of  their  diameters  or  like  sides.  The  same  remark 
is  applicable  to  frustums  of  a  cone  or  pyramid  when  the 
altitude  is  the  same,  and  the  ends  proportional. 


MENSURATION    OF    SOLIDS.  133 

PROBLEM  I. 

To  find  the  area  of  the  surface  of  a  cube. 

Rule. — Multiply  the  square  of  the  length  of  one  side  by 
the  number  of  sides,  and  the  product  will  be  the  area  of  the 
burface. 

EXAMPLES. 

1.  The  side  of  a  cube  is  18  inches  ;  hmmhh^ 
••vhat  is  the  area  of  its  surface  ?  ^HHUBir 

Here  (18^  X  6)  =  324  X  6  =  1944  in.  ■!' 

=  13.5  sq.  ft.  n||||.: 

2.  The  side  of  a  cube  is  25  inches;  e^Hjt,     -p' 
what  is  the  area  of  its  surface  ?                           x^:!^ . 

^ns,  26jV  ft.  ' 

3.  The  side  of  a  cube  is  12  feet ;  what  is  the  area  of  its 
surface  ?  ^ns.  864  ft. 

4.  The  side  of  a  cube  is  16  feet ;  what  is  the  area  of  its 
surface  ?  Jlns.  1536  ft. 

5.  The  side  of  a  cube  is  19  feet ;  what  is  the  area  of  its 
surface  ?  ^ns.  2166  ft. 

6.  The  side  of  a  cube  is  lOj  inches  ;  what  is  the  area  of 
its  surface  ?  jins.  4A|  ft. 

PROBLEM  n. 

The  area  of  the  surface  of  a  cube  being  given,  to  find  the 
length  of  the  side. 

Rule. — ^Divide  the  area  by  6,  and  extract  the  square  root 
of  the  quotient. 

EXAMPLES. 

1 .  The  area  of  a  cube  is  2400  square  inches  ;  what  is  ths 
length  of  the  side? 

Here  v/(2400  -f-  6)  ==  v/400  =  20  inches. 

2.  The  area  of  a  cube  is  24  square  feet ;  what  is  the  length 
of  the  side  ?  ^ns.  2  ft. 

12 


134  MENSURATION    OF    SOLIDS. 

3.  The  area  of  a  cube  is  216  square  feet;  what  is  the 
length  of  the  side  ?  ^ns.  6  ft. 

4    The  area  of  a  cube  is  96  feet ;  what  is  the  leufrth  of 
the  side  ?  £tis.  4  ft. 

5.  The  area  of  a  cube  is  600  square  inches  ;  what  is  the 
length  of  the  side  ?  Ans.  10  in. 

6.  The  area  of  a  cube  is  5400  square  inches  ;  what  is  the 
length  of  the  side  ?  Ans.  2$  ft. 


PROBLEM  III. 

To  find  the  solidity  of  a  cube,  the  length  of  one  of  its 
sides  being  given. 

Rule. — Multiply  the  side  by  itself,  and  that  product  again 
by  the  side,  or  cube  the  given  side,  and  it  will  give  the 
solidity  required. 

EXAMPLES. 

1.  The  side  A  B  or  B  C  of  the  cube  A  B  C  " 
G  H  E,  is  25.5  inches  ;  what  is  the  solidity  ? 

Here  (ABxBC)xAE=  (25.5  x 
25.5)  X  25.5  =  650.25  x  25.5=  16581.375 
cubic  inches. 

Or,  (25.5)''=25.5x25.5x25.5=16581.375 
cubic  inches. 

2.  What  is  the  solidity  of  a  cube  whose  side  is  5  feet  f 

Ans.  125  ft. 

3.  The  side  of  a  cube  is  15  inches  ;  what  is  its  solidity? 

Jins.  1.U53I  ft. 

4.  What  is  the  sohdity  of  a  cube  whose  side  is  5  fp.'t  3 
inches  ?  Ann.  144*|  ft. 

5.  How  many  solid  feet  are  contained,  in  a  cubic  box  whose 
depth  is  32  inches  ?  Ans.  18|f  ft. 

6.  How  many  cubic  inches  are  contained  in  a  cubic  piece 
of  timber  whose  length  is  42  inches  ?  Ans.  74088  in. 


MENSURATION    OF   SOLIDS.  135 

PROBLEM  IV. 
To  find  the  side  of  a  cube,  the  soUdity  being  given. 
Rule. — Extract  the  cube  root  of  the  solidity. 

EXAMPLES. 

1.  What  is  the  length  of  the  side  of  a  cube  containing  36 
?  -lid  feet  ? 

Here  -^36  =  3.3019,  the  side  required. 

2.  What  is  the  length  of  the  side  of  a  cube  containing 
274  cubic  inches  ?  Ans.  6.4950  in. 

3.  What  is  the  length  of  the  side  of  a  cube  whose  solidity 
is  1800  cubic  inches  1  Ans.  12.1644  in. 

4.  What  is  the  length  of  the  side  of  a  cube  whose  solidity 
is  789  cubic  feet  ?  Ans.  9.fM04  ft. 

^ROBLEM  V. 

To  find  the  solidity  of  a  parallelopipedon. 

Rule. — Multiply  the  length  by  the  breadth,  and  that  pro- 
duct again  by  the  depth,  or  altitude,  and  it  will  give  the 
solidity  required. 

EXAMPLES. 

H  G 

1 .  Required  the  solidity  of  the  paral-  bbh^^^^mi^ 
lelopipedon  A  B  C  G  H  £,  whose  ^^^^^^^^^B^ 
length  A  B  is  9  feet,  its  breadth  A  E  ^^^^^^^^^H 

5^  feet,  and    its   depth,  or   altitude,  e^^^^^H^B^Wi 

A  D  71  feet.  ^^■If ''^"'-''■'' 

Here   (ABxAExAD)=  "^ b 

(9  X  5.5)  X  7.75  =  49.5  x  7.75  = 
3S;J.625  solid  feet. 

2.  The  length  of  a  parallelopipedon  is  8  feet.  Us  breadth 
3  feet,  arid  thickness  2  feet ;  how  many  solid  feet  does  it 
tonlain  ?  Arts.  48  ft. 


136  MENSURATION    OF   SOLmS. 

3.  The  length  of  a  parallelopipedon  is  36  inches,  the 
width  20  inches,  and  depth  18  inches  ;  how  many  solid  feet 
does  it  contain  ?  Arts.  7.5  feet. 

4.  The  length  of  a  parallelopipedon  is  15  feet,  and  each 
side  of  its  square  base  31  inches  ;  what  is  the  solidity  ? 

Ans.  45.9375  ft. 

5.  What  is  the  solidity  of  a  block  of  marble,  whose  length 
is  12  feet,  breadth  5|  feet,  and  depth  2^  feet  ? 

Arts.  172^  ft. 

PROBLEM  VI. 

To  find  the  solidity  of  a  prism. 

Rule. — Multiply  the  area  of  the  base  by  the  perpen- 
dicular height  of  a  prism,  and  the  product  will  be  the 
solidity. 

EXAMPLES. 

1.  What  is  the  solidity  of  the  triangular 
prism  A  B  C  D  E  F,  whose  length  A  B  is 
20  feet,  and  either  of  the  equal  sides  B  C, 
C  F,  or  F  B,  of  one  of  its  equilateral  ends 
B  C  F,  5  feet  ? 

Here,  by  Problem  XL,  page  82,  the  area  of 
the  base  B  C  F  is  =  5'»  x  .433013  =  25  X 
.433013  =  10.825325. 

And  consequently,  10.825325  x  20  = 
216.5065  feet,  the  sohdity  required. 

2.  What  is  the  solidity  of  a  triangular  prism  whose  length 
is  18  feet,  and  one  side  of  the  equilateral  end  I5  feet? 

Ans.  17.5370  ft. 

3.  What  is  the  solidity  of  a  triangular  prism  whose  length 
is  40  feet,  and  one  side  of  the  equilateral  end  18  inches  ? 

Ans.  38.9711  ft. 

4.  What  is  the  solidity  of  a  triangular  prism  whose  lenglh 
is  24  feet,  and  one  side  of  the  equilateral  end  16  inches  ? 

Ans.  18.4752  ft. 

5.  What  is  the  value  of  a  prism  whose  height  is  32  fe.  *, 
and  each  side  of  the  equilateral  end  14  inches,  at  20  cet  ^ 
per  solid  foot  ?  Ans.  $3,772. 


MENSURATION    OF    SOLIDS. 


137 


6.  Required  the  solidity  of  a  prism  whose  base  is  a  hexa- 
gon, supposing  each  of  the  equal  sides  to  be  1  foot  6  inches, 
and  the  length  of  the  prism  10  feet.         Ans.  93.5307  ft. 


PROBLEM  VII. 

To  find  the  convex  surface  of  a  cylinder. 

Rule. — Multiply  the  circumference  or  periphery  of  the 
base,  by  the  height  of  the  cylinder,  and  the  product  will  be 
the  convex  surface  required  ;  to  which  add  the  area  of  each 
end,  and  the  sum  wih  be  the  whole  surface  of  the  cylinder 


EXAMPLES. 

1.  What  is  the  convex  surface  of  the 
right  cylinder  A  BC  D,  whose  length,  BC, 
is  24  feet,  and  the  diameter  of  its  base,  A  B, 
16  feet  ? 

Here  3.1416  x  16  =  50.2656,  the  circum- 
ference of  the  base. 

And  50.2656  x  24  =  1206.3744  square 
feet,  the  convex  surface  required. 


2.  What  is  the  whole  surface  of  a  right  cylinder,  the 
diameter  of  whose  base  is  2^  feet,  and  the  height  5  feet  ? 

Here  3.1416  x  2.5=  7.854,  the  circumference  of  the  base. 

And  7.8.54  X  5  =  39.27  square  feet,  the  convex  surface. 

Then  (2..5«x.7854) x2  =  (6.25 x. 7854) x2  =  4.90875x2 
=  9.8175  square  feet,  the  area  of  the  ends. 

Whence  39.27+9.8175  =  49.0875  square  feet,  the  whole 
surface. 

3.  Required  the  convex  surface  of  a  right  cylinder,  whose 
circumference  is  8  feet  4  inches,  and  its  length  18  feet. 

Ans.   150  ft. 

4.  What  is  the  convex  surface  of  a  right  cylinder,  the 
diameter  of  whose  base  is  2  feet,  and  its  length  30  feet  ? 

Ans.  188.496  ft. 
12* 


138 


MENSURATION    OF    SOLIDS. 


5.  What  is  the  whole  surface  of  a  right  cylinder,  the 
diameter  of  whose  base  is  lii  inches  and  its  height  20  feet  ? 

Jins.  H(;.5GS5  ft. 

6.  How  many  square  yards  are  contained  in  the  whole 
surface  of  a  cylinder,  the  diameter  of  whose  base  is  4  feet 
and  its  length  10  feet  ? 

c  Ans.  16.7552  yds. 


PROBLEM  Vin. 


To  find  the  solidity  of  a  cylinder. 

Rule. — Multiply  the  area  of  the  base  by  the  perpendicu 
lar  height  of  the  cylinder,  and  tlie  product  will  be  the 
solidity. 


EXAMPLES. 

1.  What  is  the  solidity  of  the  cylinder 
A  B  C  D,  the  diameter  of  whose  base,  A  B, 
is  30  inches,  and  the  height,  BC,  55 
inches? 

Here  (30^  X  .7854)  x  55  =  (900  X.785 1) 
X  55 =706.86x55  =38877.3  cubic  inches. 


2.  What  is  the  solidity  of  a  cylinder,  whose  height  is 
30  feet,  and  the  circumference  of  its  base  20  feet  ? 

Here     (20^  x  .07958)  x  30  =  (400  x  .07958)  x  30 
=  31.832  X  30  =  954.96  cubic  feet. 

3.  What  is  the  solidity  of  a  cylinder  whose   height  ij» 
4  feet,  and  the  diameter  of  its  base  10  inches  ? 

Ans.  2.1816  ft. 

4    The  diameter  of  a  cylinder  is  16  inches,  and  the  lengtl 
20  feet ;  what  is  the  solidity  ? 

Ans.  27.9253  ft. 

5.  The  circumference  of  a  cylinder  is  2  feet,  and  the 
length  5  feet ;  what  is  the  solidity  ? 

Ans.   1.5910  ft. 


MENSURATION    OF    SOLIDS. 


139 


0.  The  circumference  of  a  cylinder  is  20  feet,  and  the 
height  19.318  feet ;  what  is  the  solidity  ? 

Ans.  614.9305  ft. 


PROBLEM  IX.  i 

To  find  the  curve  surface  of  a  cylindric  ungula,  when 
the  section  passes  obliquely  through  the  sides  of  the  cylinder. 

Rule. — Multiply  the  circumference  of  the  base  by  half 
the  sum  of  the  greatest  and  the  least  heights  of  the  ungula, 
and  the  product  will  be  the  curve  surface. 

EXAMPLES. 

1.  What  is  the  curve  surface  of  a  cylin- 
dric ungula,  the  diameter  of  whose  base  is 
A  B,  16  feet,  and  the  greatest  and  least 
heights  are  B  F,  17,  and  A  E,  14  feet  ? 

Here  3.1416  x  16  =  50.2656,  the  cir- 
cumference of  the  base. 

Then  50.2656  x  (17  +  14)  ^  2  = 
50.2656  X  (31  -4-  2)  =50.2656  x  15.5  = 
779.1168  feet,  the  curve  surface  required. 

2.  What  is  the  curve  surface  of  a  cylindric  ungula,  the 
circumference  of  whose  base  is  21  feet,  and  the  greatest  and 
least  heights  are  13  and  8  feet?  Ans.  220.5  ft. 

3.  What  is  the  curve  surface  of  a  cylindric  ungula,  the 
diameter  of  whose  base  is  19  feet,  and  the  greatest  and  least 
heights  are  13|  and  lU  feet?  Ans.  746.13  ft. 


PROBLEM  X. 

To  find  the  solidity  of  a  cylindric  ungula,  when  the  sec- 
tion passes  obliquely  through  the  opposite  sides  of  the  cylin 
der. 

Rule. — Multiply  the  area  of  the  base  of  the  cylinder  by 
half  the  sum  of  the  greatest  and  least  heights  of  the  unguia, 
and  the  product  will  be  the  solidity. 


140  MENSURATION    OF   SOLIDS. 

EXAMPLES. 

1.  What  is  the  solidity  of  a  cylindric  ungula,  the  diame- 
ter of  whose  base,  A  B,  is  12  feet,  and  the  greatest  and  least 
heights  are  B  F,  6,  and  A  E,  4  feet  ? 

Here  (12^x.7854)  =  144x.7854  =  113.0976  square  feet, 
the  area  of  the  base. 

Then  113.0976  x  (6  +  4)  -^  2  =  113.0976  x  (10  -r-  2) 
=  1 13.0976  X  5  =  565.488  feet,  the  solidity  required. 

2.  What  is  the  solidity  of  a  cylindric  ungula,  the  diameter 
of  the  base  of  which  is  10  feet,  and  the  greatest  and  least 
heights  are  4  and  3  feet  ?  Ans.  274-89  ft. 

3.  What  is  the  solidity  of  a  cylindric  ungula,  the  circum- 
ference of  the  base  of  which  is  24  feet,  and  the  greatest  and 
least  heights  18  and  12  feet?  Ans.  687.5712  feet. 


PROBLEM  XI. 

To  find  the  convex  superficies  of  a  cylindric  rmg. 

Rule. — To  the  thickness  of  the  ring  add  the  inner  dia- 
meter, and  this  sum  being  multiplied  by  the  thickness,  and 
the  product  again  by  9.8696  (or  the  square  of  3.1416)  will- 
give  the  superficies  required. 

EXAMPLES. 

1.  The  thickness,  Ac,  of  a  cylindric 
ring  is  3  inches,  and  the  inner  diameter, 
cd,  12  inches  ;  what  is  the  convex  su- 
perficies ? 

Hei'e  [(12  +  3)  x  3]  x  9.8696  = 
(15  X  3)  X  9.8696  =  45  x  9.8696  = 
444.132  square  inches. 

2.  The  thickness  of  a  cylindric  ring  is  4  inches,  and  the 
inner  diameter  18  inches  ;  what  is  the  convex  superficies  ? 

Jlns.  868.5248  inches. 

3.  The  thickness  of  a  cylindric  ring  is  2  inches,  and  the 
mner  diameter  1  loot  6  inches;  what  is  the  convex  super- 
ficies? Jins.  394.784  inches    . 


MENSURATION    OF    SOLIDS.  141 

4.  The  thickness  of  a  cylindric  ring  is  3  inches,  and  its 
inner  diameter  9  inches  ;  what  is  the  convex  superficies  ? 

^ns.  355.3056  inches. 

5.  The  thickness  of  a  cylindric  ring  is  2  inches,  and  the 
inner  vliameter  12  inches  ;  what  is  the  convex  superficies? 

Ans.  276.3488  inches. 

6.  The  thickness  of  a  cylindric  ring  is  3.5  inches,  and  its 
inner  diameter  18.765  feet;  what  is  its  convex  superficies  ? 

Ans,  7899.4304  inches. 


PROBLEM  XII. 

To  find  the  solidity  of  a  cylindric  ring. 

Rule. — To  the  thickness  of  the  ring  add  the  inner  diame- 
ter, and  this  sum  being  multiplied  by  the  square  of  half  the 
thickness,  and  the  product  again  Jby  9.8696,  will  give  the 
sohdity. 

EXAMPLES. 

1.  What  is  the  solidity  of  an  anchor  ring,  whose  inner 
diameter  is  8  inches,  and  thickness  in  metal  3  inches  ? 

Here  [(8  +  3)  x  1.5^]  x  9.8G96  =(11  X  2.25)  x  9.8696 
=  24.75  X  9.8696  =  244.2726  cubic  inches. 

2.  What  is  the  solidity  of  an  anchor  ring  whose  inner  dia- 
meter is  9  inches,  and  the  thickness  of  metal  3  inches  ? 

Alls.  266.4792  inches. 

3.  The  inner  diameter  of  a  cylindric  ring  is  12  inches, 
and  its  thickness  4  inches  ;  what  is  its  solidity  ? 

Ans.  631.6544  inchts. 

4.  Required  the  solidity  of  a  cylindric  ring  whose  thick- 
ness is  2  inches,  and  its  inner  diameter  16  inches. 

Ans.  177.6528  inches. 

5.  Required  the  sohdity  of  a  cylindric  ring  whose  inner 
diameter  is  12  inches,  and  thickness  5  inches. 

Ans.  1048.645  inches. 

6.  What  is  the  solidity  of  a  cylindric  ring  whose  thick- 
ness is  4  inches,  and  inner  diameter  16  inches  ? 

Ans.  789.568  inches. 


142  MENSURATION    OF    SOLIDS. 


PROBLEM  XIIL 

Th«i  solidity  and  thickness  of  a  cylindric  ring  being  given, 
to  fina  the  inner  diameter. 

Role. — Divide  the  solidity  by  9.8696,  and  that  quotient 
by  the  square  of  half  the  thickness,  from  which  subtract  the 
thickness,  and  the  remainder  will  be  the  inner  diameter  of 
the  ring. 

EXAMPLES. 

1.  The  thickness  of  a  cylindric  ring  is  4  inches,  and  its 
solidity  789.568  solid  inches.     What  is  its  inner  diameter? 

Here  789.568-f-9.8696  =  80,  and  80-=-22  =  80-7-4=20, 
then  20  —  4=16  inches  the  diameter. 

2.  Required  the  inner  diameter  of  a  cylindric  ring  whose 
solidity  is  138. 1744  inches,  and  thickness  2  inches. 

Ans.  12  inches. 

3.  What  is  the  mner  diameter  of  a  cylindric  ring  whose 
solidity  is  1  solid  foot,  and  thickness  4  inches  ? 

Ans.  39.77  inches. 

4.  If  the  soHdity  of  a  cylindric  ring  be  4  solid  feet,  and 
the  thickness  3.5  inches,  what  is  the  inner  diameter  ? 

Alls.  18.765  feet. 

5.  What  must  be  the  inner  diameter  of  a  cylindric  ring 
whose  solidity  is  1  solid  inch,  and  thickness  \  of  an  inch? 

Ans.  25.8132  inches. 

6.  What  is  the  inner  diameter  of  a  cylindric  ring  whose 
solidity  is  244.2726  inches,  and  the  thickness  3  inches  ? 

Ans.  8  inches. 


PROBLEM  XIV. 
To  find  the  surface  of  a  right  cone  or  pyramid. 

Rule. — ^Multiply  the  circumference  or  perimeter  of  the 
\  Qse  by  the  slant  height  or  length  of  the  side  of  the  cone  or 
pyramid,  and  half  the  product  will  be  the  surface  required. 
And  if  this  be  added  to  the  area  of  the  base,  it  will  give  the 
whole  surface. 


MENSUBATION    OF    SOLIDS.  113 


EXAMPLES. 

1.  The  diameter  of  the  base  A  B,  of  a 
right  cone  C  A  B,  is  6  feet,  and  the  slant 
height  AC  or  B  C,  21  feet.  Required 
the  convex  surface  of  the  cone- 
Here  8.1416  X  6  =  18.8490  ==  the  cir- 
cumference of  the  base. 

And  (18.8496 x21)-r-2=895.841«-T-2 
=•197.9208  square  feet,  the  convex  sur- 
face required. 

2.  The  circumference  of  a  right  cone  is  10  feet,  and  the 
slant  height  12  feet.  What  is  the  whole  surface  of  the 
cone  ? 

Here  ( 10  x  12)  -r-  2  =  120  -^  2  =  60  square  feet,  the  con- 
vex surface. 

And  (10^  X  .07958)  =  100  X  .07958  =  7.958  square  feet, 
the  area  of  base. 

Whence  60  -f  7.958  =  67.958  square  feet,  the  whole  sur- 
face required. 

8.  Required  the  whole  surface  of  a  triangular  pyramid, 
each  side  of  its  base  being  5^  feet,  and  its  slant  height 
17^  feet. 

Here  5.5  X  3  =  16.5  =  the  perimeter  of  the  base. 

And  (16.5  X  17.5)  ^  2  =  288.75  -4-  2  =  144.875  square 
feet,  the  outward  surface  of  the  pyramid. 

Also  (5.5- X. 4880 18)  =  80.25  x.488018  =  13.0986 square 
feet,  the  area  of  *he  base. 

Whence  144.875  -f  18.0986  =  157.4736  square  feet,  the 
whole  surface  required. 

4.  The  slant  height  of  a  right  cone  is  20  feet,  and  the  dla 
meter  of  the  base  8  feet ;  required  the  convex  surface. 

Jins.  251.328  ft. 

5.  The  circumference  of  a  right  cone  is  27.5  feet,  and  the 
slant  height  11  feet;  required  the  convex  surface. 

c^/?.s.   151.25  ft. 

6.  The  slant  height  of  a  right  cone  is  20  feet,  and  the 
diameter  3  feet ;  what  is  the  whole  surface  ? 

.^7is.   101  3166  ft 


J44 


MENSURATION    OF    SOLIDS. 


7.  Required  the  outward  surface  of  a  triangular  pyramid, 
each  side  of  its  base  being  85  feet,  and  its  slant  height  14 
feet,  Ans.  73.5  ft. 

8.  The  circumference  of  a  right  cone  is  10  feet,  and  the 
prrpendicular  height  12  feet ;  required  the  convex  surface. 

Ans.  00.525  ft. 


PROBLEM  XV. 

To  find  the  surface  of  the  frustum  of  a  right  cone  or 
pyramid. 

Rule. — Multiply  the  sum  of  the  perimeters  of  the  two 
ends  by  the  slant  height  of  the  frustum,  and  half  the  pro- 
duct will  be  the  surface  required. 


EXAMPLES. 

1.  In  the  frustum  of  the  cone  A  B  D  E, 
the  circumferences  of  the  two  ends  A  B 
and  ED,  are  22.75  and  15.5  feet  respect- 
ively, and  the  slant  height,  A  E,  is  26  feet ; 
required  the  convex  surface. 

Here  [(22.75  +  15.5)  x  2G]  -f-  2  = 
(38.25  X  26)  -^  2  =  994.5  -^  2  =  497.25 
square  feet,  the  convex  surface  required. 

2.  Required  the  surface  of  the  frustum  of  a  square 
pyramid,  one  side  of  the  base  being  121  feet,  and  of  the 
upper  end  5|  feet,  and  its  slant  height  40|  feet. 

Here  121  x  4  =  50  =  the  perimeter  of  the  base,  and 
5|  X  4  =  23  =  the  perimeter  of  the  upper  end. 

Then  [(50  -f  23)  x  40.25]  -^  2  =  (73  x  40.25)  -h  2  = 
2938.25 -T-  2  =  1469.125  square  feet,  the  surface  required. 

3.  What  is  the  convex  surface  of  the  frustum  of  a  right 
cone,  the  circumference  of  the  greater  end  being  23|  feet, 
and  that  of  the  less  end  16|  feet,  and  the  length  of  the  slant 
side  12  feet  ?  Ans.  240  ft. 

4.  What  IS  the  convex  surface  of  the  frustum  of  a  right 
cone,  the  diameters  of  the  ends  being  8  and  4  feet,  and  the 
length  of  the  slant  side  20  feet  ?  Ans.  376.992  ft. 


MENSURATION    OF    SOLIDS. 


145 


5.  Required  the  surface  of  a  hexagonal  pyranaid,  one  side 
of  the  base  being  85  feet,  and  of  the  upper  end  3f  feet,  and 
the  slant  height  20|  feet.  Ans.  733.5  ft. 

6.  What  is  the  convex  surface  of  the  frustum  of  a  right 
cone,  the  diameters  of  the  ends  being  5  and  4  feet,  and  the 
slant  height  6  feet  ?  Ans.  84.8232  ft. 


PROBLEM  XVI. 

To  find  the  solidity  of  a  cone  or  pyramid. 

Rule. — ^Multiply  the  area  of  the  base  by  the  perpendicu- 
lar height  of  the  cone  or  pyramid,  and  one-third  of  the  pro- 
duct will  be  the  solidity. 


EXAMPLES. 

1.  Required  the  solidity  of  a  cone  CAB, 
whose  diameter,  A  B,  is  30  feet,  and  its  per- 
pendicular height,  G  C,  36  feet. 

Here  (.7854  X  30^)  =  .7854  X  900  = 
706.86  =  the  area  of  the  base. 

And  (706.86  x  36)  -=-  3  =  25446.96  -i-  3 
=  8482.32  feet,  the  solidity. 


2.  Required  the  solidity  of  the  hex- 
agonal pyramid  H  A  D  H,  each  of  the 
equal  sides  of  its  base  being  20  feet,  and 
the  perpendicular  height,  H  G,  50  feet. 

Here,  by  the  table,  page  82,  for  poly- 
gons we  have  2.598076  (multipKer  when 
the  side  is  1)  x  20^  =  2.598076  x  400  = 
1039.2304  =  the  area  of  the  base. 

And  (1039.2304  x  50)^ 3  =  51961.52 
-*-  3  =  17320.5066  feet,  the  solidity. 


3.  What  is  the  solidity  of  a  cone,  the  diameter  of  whose 
hone  is  18  inches,  and  its  altitude  24  feet  ? 

Ans.  14.1373  ft. 
13 


146  MENSURATION    OF    SOLIDS. 

4.  If  the  circumference  of  the  base  of  a  cone  be  60  feet; 
and  its  height  72  feet ;  what  is  the  solidity  ? 

^ns.  6875.712  ft. 

5.  What  is  the  solidity  of  a  triangular  pyramid,  whose 
height  is  30  feet,  and  each  side  of  the  base  3  ft.  ? 

Jns.  38.9711  ft. 

6.  yVhat  is  the  solidity  of  a  pentagonal  pyramid,  its  height 
being  12  feet,  and  each  side  of  its  base  2  feet  ? 

^ns.  27.5276  ft. 

7.  What  is  the  solidity  of  a  cone,  whose  diameter  of  the 
base  is  14  feet,  and  the  slant  side  being  25  feet  ? 

Ans.  1231.5072  ft. 


PROBLEM  XVII. 

To  find  the  solidity  of  the  frustum  of  a  cone  or  pyramid. 

1.  For  the  frustum  of  a  cone,  the  diameters  of  the  two 
ends  and  the  height  being  given. 

Rule. — ^Divide  the  difference  of  the  cubes  of  the  diameters 
of  the  two  ends  by  the  difference  of  the  diameters,  and  this 
quotient  being  multiplied  by  .7854,  and  again  by  one-third 
of  the  height,  will  give  the  solidity. 

Or  divide  the  difference  of  the  cubes  of  the  circumferences 
of  the  two  ends,  by  the  difference  of  the  circumferences,  and 
the  quotient  being  multiplied  by  .07958,  and  again  by  one- 
third  of  the  height,  will  give  the  solidity. 

Or  to  the  product  of  the  diameters  add  one-third  of  the 
square  of  their  difference,  and  that  sum  being  multiplied  by 
.7854,  and  again  by  the  height,  will  give  the  solidity. 

Or  to  the  product  of  the  circumferences  add  one-third  of 
the  square  of  their  difference,  and  this  sum  being  multiplied 
by  .07958,  and  again  by  the  height,  will  give  the  sohdity. 

2.  For  the  frustum  of  a  pyramid,  the  sides  of  the  base 
and  the  height  being  given. 

Rule. — To  the  areas  of  the  two  ends  of  the  frustum  add 
the  square  root  of  their  product,  and  this  sum  being  multi- 
plied by  one-third  of  the  height,  will  give  the  solidity. 


MENSURATION    OF    SOLIDS. 


147 


EXAMPLES. 

1.  What  is  tlie  solidity  of  the  frustum 
of  the  cone  A  B  D  E,  the  diameter  of  whose 
greater  end,  A  B,  is  6  feet,  that  of  the  less 
end,  E  D,  4  feet,  and  the  perpendicular 
heiaht,  F  G,  9  feet  ? 

Here  (6''  —  4^)  -=-(6  —  4)  =  (216  —  64) 
-T-  2  =  152  ^  2  =  76. 

And  (76  X  .7854)  x  (9  -^  3)  =  59.6904 
X3  =  179.0712  feet,  the  solidity  required. 

Or  [(6  X  4)  4-  (6  -  4)  ^^  3]  =  24  +  (2"-^  3)  =  (24+-*) 
=251.  And  (25|  X.7854)  x  9  =  19.8968  x  9=179.0712 
feet,  the  solidity,  as  before. 

2.  What  is  the  solidity  of  the  frustum  of  a  cone,  the  cir- 
cumference of  the  greater  end  being  40  feet,  and  that  of  the 
less  20  feet,  and  the  length  or  height  51  feet  ? 

Here  (40»  —  20'')  -t-  (40  —  20)  =  (64000  —  8000)  —  20 
=  56000  -4-  20  =  2800. 

And  (2800  X  .07958)  x  (51  ^  3)  =  222.824  x  17 
=  378S.008  feet,  the  solidity  required. 

Or  [(40  X  20)  +  (40  -  20)  ^--3]  =800  +  (20«-^)=800 
+  133^  =  93:)^,  and  (933^  x  .07958)  x  51  =  74.274666 
X  51  =3788.008  feet,  the  solidity,  as  before. 

3.  What  is  the  solidity  of  the  frustum  aADd  of  an 
hexagonal  pyramid,  the  side  A  B  of  whose  greater  end  is 
4  feet,  that  ah  of  the  less  end  3,  and  the  height  G^  9  feet  ? 

Here  2.598076  (the  tabular  multiplier) 
X  33  =  2.598076  X  9  =  23.382684  the 
area  of  the  less  end. 

And  2.598076  x  4^  =  2.598076  x  16 
=  41.509216  the  area  of  the  greater 
end. 

Whence  ^Z (23. 382684x4 1.5692 16) 
=  ^/971,999841  =  31.176912. 

And  (23.382G84  + 41.569216 +31. 
176912)  X  (9  -T-  3)  =  96.128812  x  3  =  288.3864.36  feet. 

4.  What  is  the  solidity  of  the  frustum  of  a  cone,  the  dia- 
meter of  the  greater  end  being  4  feet,  that  of  the  less  end  2, 
und  the  altitude  9  feet  ?  ^ns.  65.9736  feet 


148  MENSURATION    OF    SOLIDS. 

5.  What  is  the  solidity  of  the  frustum  of  a  cone,  the  dia- 
meter of  the  greater  end  of  which  is  5  feet,  that  of  the  less 
end  3  feet,  and  the  ahitude  4  feet  ?       Ans.  51.3128  feet. 

6.  What  is  the  solidity  of  the  frustum  of  a  cone,  the  cir- 
cumference of  the  greater  end  being  20  feet,  that  of  the  less 
end  10  feet,  and  the  length  or  height  21  feet  ? 

Ans.  389.942  feet. 

7.  What  is  the  solidity  of  the  frustum  of  a  cone,  the  cir- 
cumference of  the  greater  end  being  12  feet,  that  of  the  less 
end  8  feet,  and  the  height  5  feet  ?         Ans.  40.3205  feet. 

8.  What  is  the  solidity  of  the  frustum  of  a  square  pyra- 
mid, one  side  of  the  greater  end  being  18  inches,  that  of  the 
less  end  15  inches,  and  the  altitude  60  inches  ? 

Ans.  9.4791  feet. 

9.  What  is  the  solidity  of  the  frustum  of  an  equilateral 
triangular  pyramid,  one  side  of  the  greater  end  being  14 
inches,  that  of  the  less  end  8  inches,  and  the  height  10  feet? 

Ans.  3.7287  feet. 

10.  What  is  the  solidity  of  the  frustum  of  a  pentagonal 
pyramid,  the  side  of  whose  greater  end  is  18  inches,  that  of 
the  less  end  12  inches,  and  the  height  4  feet  6  inches  ? 

Ans.  12.2583  feet 


PROBLEM  XVIII. 

The  solidity  and  altitude  of  a  cone  being  given,  to  /  nd 
the  diameter. 

Rule. — Divide  the  solidity  by  the  product  of  .7854,  nd 
one-third  of  the  altitude,  and  the  square  root  of  the  quel  .^nt 
will  be  the  diameter. 

EXAMPLES. 

1.  The  solidity  of  a  cone  is  16  feet,  and  the  altitui  »  9 
feet ;  what  is  the  diameter  ? 

Here  v/{16  -h  [.7854  x  (9 -=-3)]}  =  ^/[16-f-(.7854x.<)] 
=V'(16^2.3562)=-v/6.7906=2.6057  feet,  the  diameter. 

2.  The  altitude  of  a  cone  is  15  feet,  and  the  solidity  3P 
feet ;  what  is  the  diameter  ?  Ans,  2.7639  ft  ^ 


MENSURATION    OF    SOLIDS.  149 

3.  The  solidity  of  a  cone  is  18  feet,  and  the  altitude  9 
feet;  what  is  the  diameter?  Ans.  2.9316  feet. 

PROBLEM  XIX. 

The  solidity  and  diameter  of  a  cone  being  given,  to  find 
the  altitude. 

Rule. — Divide  the  solidity  hy  the  product  of  .7854  and 
the  square  of  the  diameter,  and  the  quotient,  being  multiplied 
by  3,  will  give  the  altitude. 

EXAMPLES. 

1.  The  solidity  of  a  cone  is  30  feet,  and  the  diameter 
2  feet ;  what  is  the  altitude  ? 

Here  [30  -=-  (.7854  x  2^]  X  3  =  [30  -.  (.7854  X  4)]  x  3 
=  (30  -5-  3.141G)  X  3  =  9.5492  x  3  =  28.6476  feet,  the  al- 
titude. 

2.  The  diameter  of  a  cone  is  20  inches  ;  what  must  be 
the  altitude,  to  make  20  solid  feet  ?  Ans.  27.5019  ft. 

3.  The  solidity  of  a  cone  is  2513.28  feet,  and  the  diame- 
ter 20  feet ;  what  is  the  altitude  ?  Ans.  24  ft. 

PROBLEM  XX. 

The  altitude  of  a  cone  or  pyramid  being  given,  to  divide 
it  into  two  or  more  equal  parts,  by  sections  parallel  to  the 
base,  to  find  the  perpendicular  height  of  each  part. 

Rule. — Multiply  the  cube  of  the  altitude  by  the  numera- 
tor of  the  proportion  left  at  the  vertex,  and  divide  the  pro 
duct  by  the  denominator ;  the  cube  root  of  the  quotient  will 
be  the  altitude  of  the  cone  or  pyramid  left  at  the  vertex. 


EXAMPLES. 

1 .  The  altitude  of  the  cone  A  B  C  is 
10  feet,  to  be  divided  into  three  equal 
parts  by  sections  parallel  to  the  base ; 
required  the  perpendicular  height  of  each 
part. 

Here  ^/[(10«X  1) -T-3]  =  ^3/[(1000xl) 
-5-3]=  v^/(l(X)0  -^  3)  =  ^3/.333j  =  6.9336 
<eet,  the  altitude  of  the  first  section  =  C  H. 
13* 


150  MENSURATION    OF    SOLIDS. 

And  ^/  [(10'x2)-5-  8]  =  ^[(1000  x  2)  -r-  3]  =  ^(2000 
-T-  3)  =  ^  666f  =  8.7358  feet,  the  altitude  C  I.  Now 
C  I  —  C  H  ==  8.7358  —  6.9336  =  1.8022  feet,  the  altitude 
of  the  second  section  I  H. 

Then  C  K  -  C  I  =  10  —  8.7358  =  1.2642  feet,  the  alti- 
tude of  the  third  section  I K. 

2.  The  altitude  of  a  pyramid  is  12  feet,  to  be  divided  into 
three  equal  parts  by  sections  parallel  to  the  base ;  required 
the  perpendicular  height  of  each  part. 

^ns.  8.3203,  2.1626,  and 

1.5171  feet,  the  altitudes. 

3.  The  altitude  of  a  cone  is  20  feet,  to  be  divided  into 
four  equal  parts  by  sections  parallel  to  the  base ;  required 
the  perpendicular  height  of  each  part. 

Jlns.  12.5992,  3.2748,  2.2972,  and 
1.8288  feet,  the  altitudes. 


PROBLEM  XXI. 

To  find  the  solidity  of  an  ungula  when  the  section  passes 
through  the  opposite  extremities  of  the  ends  of  the  frustum. 

Rule. — From  the  square  of  the  greater  diameter  subtract 
the  square  root  of  the  product  of  the  two  diameters,  multi- 
plied by  the  less  diameter. 

This  difference  being  divided  by  the  difference  of  the 
diameters,  and  the  quotient,  multiplied  by  the  greater  diame- 
ter, that  product  by  the  height,  and  the  last  product  by  .2618 
will  give  the  solidity. 

EXAMPLES. 

1.  Required  the  solidity  of  a  conical  ,^^~~^^a 
ungula,  the  diameter  of  the  greater  end  /  J^^^L 
being  5  feet,  that  of  the  less  end  1 .8  feet,       /  ^^^^^ 

(i.8  X  3)  -J-  3.2]'=  [(25 -5.4) -=-3.2]=  '^^j!^^^^ 
(19  6 -i- 3.2)  =»  6.125. 


MENSURATION    OF    SOLIDS. 


151 


And  (6.125  X  5  X  12  X  .2618) 
lidity. 


96.2115  feet,  the  so- 


2.  Required  the  solidity  of  a  conical  ungula,  the  diametei 
of  the  greater  end  being  10  feet,  that  of  the  less  end  21  feet, 
and  the  height  15  feet.  Ans.  458.15  ft. 

3.  Required  the  solidity  of  a  conical  ungula  the  diameter  of 
the  greater  end  being  4.23  inches,  that  of  the  less  end  3.7 
inches,  and  the  height  5.7  inches. 

Ans.  38.7692  in. 


PROBLEM  XXII. 

To  find  the  soUdity  of  a  cuneus  or  wedge. 

Rule. — Add  twice  the  length  of  the  base  to  the  length  of 
the  edge,  then  multiply  this  sum  by  the  height  of  the  wedge, 
and  again  by  the  breadth  of  the  base,  and  one  sixth  of  the 
last  product  will  be  the  solidity. 

EXAMPLES. 

1.  How  many  solid  feet  are  there  in 
a  wedge  whose  base  is  5  feet  4  inches 
long,  and  9  inches  broad,  the  length  of 
the  edge  being  3  feet  6  inches,  and  the 
perpendicular  height*2  feet  4  inches  ? 

Here  5  ft.  4  in.  =  64  in.,  3  feet  6  in. 
=  42  in.  and  2  ft.  4  in.  =  28  in. 

Then  (64 x24-42)x28  =(128+42) 
X  28  =  170  X  28  =  4760. 

And  (4760  x  9)  -^  6  =  42840 


C  B 

6  =  7140  solid  inches. 
Whence  7140  -r-  1728  =  4.1319  solid  feet,   ' 


2.  The  length  and  breadth  of  the  base  of  a  wedge  are 
35  and  15  inches,  the  length  of  the  edge  55  inches, 
and  the  perpendicular  height  18  inches;  what  is  the 
solidity  ?  Jlns.  3.2552  ft. 

3.  The  length  and  breadth  of  the  base  of  a  wedge  are 
27  and  8  inches,  the  length  of  the  edge  36  inches,  and 
the  perpendicular  height  3  feet  6  inches ;  what  is  the 
solidity?  dns.  2.9166  ft. 


152  MENSURATION    OF   SOLIDS. 

PROBLEM  XXm. 
To  find  the  solidity  of  a  prismoid. 

Rule. — Multiply  the  sum  of  the  lengths  of  the  two  ends 
by  the  sum  of  the  breadths,  and  add  this  product  to  the  sum 
of  the  areas  of  the  two  ends ;  multiply  the  result  by  one 
sixth  of  the  height,  and  the  product  will  be  the  solidity. 

EXAMPLES. 

1.  What  is  the  solidity  of  a  rectan- 
gular prismoid,  the  length  and  breadth 
of  one  end  being  14  and  12  inches,  and 
the  corresponding  sides  of  the  other  6 
and  4  inches ;  and  the  perpendicular 
30a  feet? 

Here  (by  Prob.  I,  p.  01)  E  F  x  E  H  dI 
=  6  X  4  =  24  square  inches,  the  area 
of  the  less  end,  E  F  G  H. 

And,  by  the  same  problem  ABxAD  =  14xI2 
square  inches,  the  area  of  the  greater  end,  A  B  C  D. 

Also  305ft.  =  366  inches,    calling  the  height  h, 

[(A  B  +  E  F)  X  (A  D  +  E  H)  +  (E  F  G  H   + 

ABCD)]x^  =  [(14  +  6)x(12-f  4)  +  (168  +  24)]'^. 

=  [(20  X  16)  +  192]  X  61  =  (320+192)  X  61  =  512  x  61 
=  31232  cubic  inches,  the  solidity  required. 

2.  What  is  the  solidity  of  a  rectangular  prismoid,  the 
length  and  breadth  of  one  end  being  12  and  8  inches,  and 
the  corresponding  sides  of  the  other  8  and  6  inches,  and  the 
perpendicular  height  5  feet?  Ans.  2.4537  ft. 

3.  What  is  the  solidity  of  a  stick  of  hewn  timber,  whose 
ends  are  respectively  30  by  27  inches,  and  24  by  18  inches, 
and  whose  length  is  48  feet  ?  Ans.  204  ft. 

4.  What  is  the  capacity  of  a  coal  wagon,  whose  inside 
dimensions  are  as  follows :  at  the  top  the  length  is  7  feet, 
and  breadth  6  feet,  at  the  bottom  the  length  is  5  feet,  and 
breadth  3  feet,  and  the  perpendicular  depth  is  4  feet? 

Ans.   110  ft 


MENSURATION    OF    SOLIDS.  163 

PROBLEM  XXIV. 
To  find  the  convex  surface  of  a  sphere. 

Rule. — ^Multiply  the  diameter  of  the  sphere  by  its  cir- 
cumference, and  the  product  will  be  the  convex  superficies 
required. 

The  curve  surface  of  any  zone  or  segment  will  also  be  found 
by  multiplying  its  height  by  the  whole  circumference  of  the 
sphere. 

EXAMPLES. 

1.  What  is  the  convex  surface  of  a 
globe  A  D  B  C,  whose  diameter,  A  B, 
is  16  inches  ? 

Here  (3.1416x16) X 16  =.50.2650 
X  16  =  804.2496  square  inches,  the 
surface  required. 

2.  What  is  the  convex  surface  of  a 
sphere  whose  diameter  is  10  feet  ? 

J2ns.  314.16  ft. 

3.  What  is  the  convex  surface  of  a  sphere  whose  diameter 
is  4  feet  ?  ^ns.  50.2656  ft. 

4.  The  diameter  of  a  globe  is  21  inches  ;  what  is  the  con- 
vex surface  of  that  segment  of  it  whose  height  is  4^  inches? 

^ns.  296.S812  inches. 

5.  What  is  the  convex  surface  of  a  sphere  whose  diame- 
ter is  6  feet  ?  Jins.   1 13.0976  ft. 

6.  If  the  diameter  of  the  globe  we  inhabit  be  793.5  miles; 
what  is  the  convex  surface  ?  Jitis.  197808409.26  miles. 

PROBLEM  XXV. 

To  find  the  solidity  of  a  sphere  or  globe. 

Rule. — Multiply  the  cube  of  the  diameter  by  .5236,  and 
the  product  will  be  the  solidity. 


154  MENSURATION   OF   SOLIDS. 


EXAMPLES. 

1.  What  is  the  solidity  of  a  gJobe  whose  diameter  is  4.5 
feet  ? 

Here  (4.5='  x  .5236)  =  4.5  x  4.5x4.5x.5236=91.125x 
.5236  =47.71305  solid  feet. 

2.  What  is  the  solidity  of  a  globe  whose  diameter  is  34 
feet?  ^ns.  22.4493ft. 

3.  What  is  the  solidity  of  a  globe  whose  diameter  is  17 
inches?  ^ns.   1.4886  ft. 

4.  What  is  the  solidity  of  a  globe  whose  diameter  is  3 
feet  4  inches  ?  ^ns.  19.3925  ft. 

5.  How  many  cubic  miles  are  contained  in  the  solidity  of 
the  earth,  if  its  diameter  be  7935  miles  ? 

^ns.  261601621246.35  miles. 


PROBLEM  XXVI. 

The  convex  surface  of  a  globe  being  given,  to  find  its  dia- 
meter. 

Rule. — Multiply  the  given  area  by  .31831,  and  the  square 
root  of  the  product  will  be  the  diameter. 

EXAMPLES. 

1.  What  is  the  diameter  of  that  globe,  the  area  of  whose 
convex  surface  is  14  square  feet  ? 

Here  v/(14  x  .31831)  =  ^4.45634  =  2.1110  feet,  the 
diameter  required. 

2.  The  convex  surface  of  a  sphere  is  one  square  rood  ; 
what  is  its  diameter  ?  ^ns.  3.5682  rods. 

3.  The  expense  of  gilding  a  ball  at  $1.80  per  square  foot 
is  thirty-four  dollars  ;  what  is  its  diameter  ? 

^ns.  2.452  ft. 


MENSURATION    OF   SOLIDS. 


155 


PROBLEM  XXVII. 
The  solidity  of  a  globe  being  given,  to  find  the  diameter. 

Rule. — Divide  the  solidity  by  .5236,  and  extract  the  cube 

root  of  the  quotient. 

EXAMPLES. 

1.  The  solidity  of  a  globe  is  2000  solid  inches  ;  what  is 
its  diameter  ? 

Here  ^( 2000 -f-. 5236)  =  ^^3819.7097  =  15.631  inches 
the  diameter. 

2.  The  solidity  of  a  globe  is  10  sohd  feet ;  what  is  its  dia- 
meter ?  Ans.  2.67  ft. 

3.  What  is  the  circumference  of  a  globe  whose  solidity 
is  8  solid  feet  ?  Ans.  7.7952  ft. 


PROBLEM  XXVm. 
To  find  the  solidity  of  the  segment  of  a  sphere. 

Rule. — To  three  times  the  square  of  the  radius  of  its  base 
add  the  square  of  its  height ;  and  this  sum  multiplied  by  the 
height,  and  the  product  again  by  .5236,  will  give  the  solidity. 

Or,  from  three  times  the  diameter  of  the  sphere  subtract 
twice  the  height  of  the  segment,  multiply  by  the  square  of 
the  height,  and  that  product  by  .5236 ;  the  last  product  will 
be  the  solidity. 


examples. 

1.  The  radius  An  of  the  base  of  the 
segment  CAB  is  7  inches,  and  the 
height  Cn  4  inches  ;  what  is  the  soli- 
dity ? 

Here  [(7«  x  3)  +  4^]x4=  [(49  x  3) 
+  16]  X  4  =  (147  +  16)  X  4  =  163 
X  4  =  652. 

Then  652  X  .5236  =  341.3872  solid 
inches 


156  MENSURATION    OF    SOLIDS. 

3.  The  diameter  of  a  sphere  is  6  inches.  What  is  the 
solidity  of  the  segment  whose  height  is  2  inches  ? 

Here  [76  x  3)  -  (2  x  2)]  x  2^  =  (18  —  4)  x  4  =  14  x  4 
=  56.     Then  56  x  .5236  =  29.3216  solid  inches. 

3.  What  is  the  solidity  of  a  spherical  segment,  the  dia- 
meter of  its  base  being  40  inches,  and  the  height  10  inches  ? 

^tis.  3.9391  feet. 

4.  The  diameter  of  a  sphere  is  18  inches  ;  what  is  the  soli- 
dity of  a  segment  cut  from  it,  the  height  being  3  inches  ? 

^ns.  226.1952  inches. 

5.  The  diameter  of  a  spherical  segment  is  20  inches,  and 
the  height  6  inches ;  how  many  gallons  of  water  will  it 
bold,  each  gallon  containing  282  cubic  inches  ? 

^ns.  3.743  gallons. 


PROBLEM  XXIX. 
To  find  the  solidity  of  a  frustum  or  zone  of  a  sphere. 

Rule.— To  the  sum  of  the  squares  of  the  radii  of  the  two 
ends,  add  one-third  of  the  square  of  their  distance,  or  of  the 
breadth  of  the  zone,  and  this  sum  multiplied  by  the  said 
breadth,  and  the  product  again  by  1.5708,  will  give  the  so- 
lidity. 

EXAMPLES. 

1.  What  is  the  solidity  of  the  zone 
A  BC  D,  whose  greater  diameter,  A  B, 
is  1  foot  8  inches,  the  less  diameter, 
D  C,  1  foot  3  inches,  and  the  distance 
nm  of  the  two  ends  10  inches  ? 

Here  [(Am^  +  Dn^')  +  ^  (nm)"] 
xnmx  1.5708  =  [(10«  +  7^3)  +  (10» 
.^3)]xl0  X  1.5708=(100  +  561+331)  x  15.7(58=189^3 
X  15.708  =  2977.975  cubic  inches  the  solidity  of  the 
zone  required. 

2.  What  is  the  solidity  of  a  zone  whose  greater  diameter 
is  9  feet  3  inches,  less  diameter  6  feet  9  inches,  and  height 
5  feet  6  inches.  ^ns.  370.3242  feet. 

3    What  is  the  solidity  of  a  zone,  whose  greater  diameter 


MENSURATION    OF    SOLIDS.  157 

is  2  feet,  the  less  diameter  1  foot  8  inches,  and  the  distance 
of  the  ends  4  inches  ?  ^ns.   156(3.(5112  inches. 

4.  Required  the  solidity  of  the  middle  zone  of  a  sphere, 
whose  top  and  bottom  diameters  are  each  3  feet,  and  the 
breadth  of  the  zone  4  feet  ?  Ans.  61.7848  feet. 


PROBLEM  XXX. 

To  find  the  solidity  of  a  circular  spindle,  its  length  and 
middle  diameter  being  given. 

Rule. — To  the  square  of  half  the  length  of  the  spindle,  or 
longest  diameter,  add  the  square  of  half  the  middle  diameter, 
and  this  sum  divided  by  the  middle  diameter  will  give  the 
radius  of  the  circle. 

2.  Take  half  the  middle  diameter  from  the  radius  thus 
found,  and  it  will  give  the  central  distance ;  or  that  part  of 
the  radius  that  lies  between  the  centre  of  the  circle,  and  that 
of  the  spindle. 

3.  Find  the  area  of  the  generating  circular  segments,  by 
Problem  XL,  rule  5,  page  104. 

4.  From  one-third  of  the  cube  of  half  the  length  of  the 
spindle,  subtract  the  product  of  the  central  distance,  and 
half  the  area  last  mentioned,  and  the  remainder  multiplied 
by  12.5664,  will  give  the  soUdity  of  the  spindle. 

EXAMPLES. 

I .  The  longest  diameter,  A  B, 
ol  the  circular  spindle  ADBC, 
is  48  inches,  and  the  middle  dia- 
meter, C  D,  36  inches ;  what  is 
the  sohdity  of  the  spindle  ?  f- J Iq 

First  (Ac2+Ce^)-=-CD=(24''  ^ 

-f  18^)  ~2Q  =  (576  +  324)  -^  36  =  900  -^  36  =  25,  the 
radius  OC. 

Second,  OC  —  Ce  =  25  —  18  =  7,  O  e,  the  central  dis- 
tance. 

Third,  (by  Problem  XL,  rule  5,  page  104,)  Cc  -^2  OC 
«=  18  -^  50  =  .36,  the  tabular  versed  sine,  against  which 
(Stands  .254550  the  tabular  segment. 

14 


158  MENSURATION    OF    SOLIDS 

Here  .254550  x  50^=. 254550  X  2500  =  636.375,   ihe 

area  of  the  segment  A  B  C  A. 

Fourth,  iAe^—  (^  ABCA  x  Oe)xl2.5664=  [(24«-=-3) 
-  (636.375-=-2)x7]x  12.5664  =  [(13824--3)— (318.1875 
X7)]  x  12.5664= (4608-2227.3125)  x  12.5664=2380.6875 
X  12.5664  =  29916.6714,  solidity  of  the  spindle. 

2.  If  the  length  of  a  circular  spindle  be  40  inches,  and  its 
middle  diameter  30  inches,  what  is  its  solidity  ? 

^ns.  17312.8886  inches. 


PROBLEM  XXXI. 

To  find  the  solidity  of  the  middle  frustum  of  a  circular 
spindle,  its  length,  the  middle  diameter,  and  that  of  either 
of  the  ends,  being  given. 

Rule  1. — Divide  the  square  of  half  the  length  of  the  frus- 
tum by  half  the  difference  of  the  middle  diameter,  and  that  of 
either  of  the  two  ends ;  and  half  this  quotient,  added  to  one- 
half  of  the  said  difference,  will  give  the  radius  of  the  circle. 

2.  Find  the  central  distance,  and  the  revolving  area,  as 
in  the  last  problem. 

3.  From  the  square  of  the  radius  take  the  square  of  the 
central  distance,  and  the  square  root  of  the  remainder  will 
give  half  the  length  of  the  spindle. 

4.  From  the  square  of  half  the  length  of  the  spindle  take 
one-third  of  the  square  of  half  the  length  of  the  frustum,  and 
multiply  the  remainder  by  the  said  half  length. 

5.  From  this  product  take  that  of  the  generating  area  and 
central  distance,  and  the  remainder  multiplied  by  6.2832 
will  give  the  solidity  of  the  fnistum. 

Examples. 


1 .  What  is  the  solidity  of  the 
middle  frustum,  A  B  C  D,  of  a 
circular  spindle,  whose  middle 
diameter,  nm,  is  36  inches,  the 
diameter  D  A  or  C  B,  of  the  end 
16  inches,  and  its  length  or  40 
inches  ? 


E,c... 


MENSURATION    OF    SOLIDS.  159 

First,  i  [Oe«  ~  (?ie  — Do)]  +  k  (ne-Do)  =  h  [20^^(1^ 
-8)]+  d  (18  -  8)  =  ^  (,40a  -r-  10)  +  (10  -i-  2)  =20+5 
=  25,  the  radius  of  the  circle. 

Second,  25  —  ne  =  25  —  18  =  7,  the  central  distance. 

And  ne  —  Do  =  18  —  b  =  10,  the  versed  sine  of  the 
arc  Drt. 

Then,  by  Problem  XL,  rule  5,  page  104,  10  ^  (25  x  2) 
=  10 -T- 50  =  .2,  the  tabular  versed  sine;  against  which 
stands  .111823  the  tabular  segment. 

Hence  .11 182:i  X  50^=.111»23  x  2500  =  279.5575,  the 
area  of  the  revolving  segment  DCnD. 

Again,  by  Problem  1,  page  01,  or  x  Do  =40x8  =  320, 
the  area  of  the  rectangle  DorCD. 

And  DC/iD  +  D  or  CD  =  279.5575  +  320  =  599.5575, 
the  generating  area  OrCnDo. 

Third,  ^/(25^  —  7-)  =  v/(625  —  49)  =  ^/576  =  24  = 
Ec,  half  the  length  of  the  spindle. 

Fourth,  {[(Ee^—  ^  Oe^)  x  Oe]  —(OrCnDox7)}X 6.2832 
=  {[(24=  —  133i)  X  20]  —  (599.5575  X  7)f  X  6.2832  = 
[(442f  X  20)  —  4196.9025]  x  6.2832  =  (8853.3333^ — 
4196.9025)  X6.2832  =  4656.4308^  x6.2832  =  29257.2862 
cubic  inches,  the  soHdity  of  the  middle  frustum  AmBCnDA 
required. 

2.  The  middle  diameter  of  the  frustum  of  a  circular  spin- 
dle is  2  feet  8  inches,  the  diameter  at  the  end  is  2  feet,  and 
the  length  3  feet  4  inches  ;  what  is  the  solidity  ? 

^ns.  27285.0882  inches. 

PROBLEM  XXXII. 

To  find  the  solidity  of  a  spheroid,  its  two  axes  being  given. 

Rule. — Multiply  the  square  of  the  revolving  axis  by  the 
fixed  axis,  and  this  product  again  by  .523(5,  or  one-sixth  of 
3.1416,  and  it  will  give  the  solidity  required. 

EXAMPLES. 

1.  In  the  prolate  spheroid  A  BCD, 
the  transverse  or  fixed  axis,  AC,  is  3  ft, 
and  the  conjugate  or  revolving  axis, 
DB,  is' 2  feet ;  what  is  the  solidity  ? 

Here  (2-  X  3)  x  .5230  =  (4  x  3^ 
X.523()  =\2x  .5-J3(}  =  ;}.-.ib32  feet, 
the  solidity  required. 


IGO  MENSURATION    OF    SOLIDS. 

2.  What  is  the  solidity  of  a  prolate  spheroid  whose  trans- 
verse or  fixed  axis  is  4  feet  2  inches,  and  conjugate  or  re- 
volving axis  3  feet  4  inches  ?  Ans,  24.2407  ft. 

3.  What  is  the  solidity  of  a  prolate  spheroid  whose  fixed 
axis  is  8  feet  4  inches,  and  its  revolving  axis  5  feet  ? 

Ans.  109.0833  ft. 

4.  What  is  the  solidity  of  an  oblate  spheroid  whose  con- 
jugate or  fixed  axis  is  5  feet,  and  its  transverse  or  revolving 
axis  8  feet  4  inches  ?  Ans.  181.8055  ft. 

5.  What  is  the  solidity  of  an  oblate  spheroid  whose  fixed 
axis  is  30  inches,  and  its  revolving  axis  40  inches  ? 

Ans.  14.5444  ft. 

PROBLEM  XXXm, 

To  find  the  solidity  of  the  middle  frustum  of  a  spheroid, 
its  length,  the  middle  diameter,  and  that  of  either  of  the  ends, 
being  given. 

Case  1. — When  the  ends  are  circular,  or  perpendicular  to 
the  fixed  axis. 

Rule. — To  twice  the  square  of  the  middle  diameter  add 
the  square  of  the  diameter  of  either  of  the  ends,  and  this  sum 
multiplied  by  the  length  of  the  frustum,  and  the  product 
again  by  .2618  (or  one-twelfth  of  3.141(5),  will  give  th« 
solidity. 

EXAMPLES. 

1.  In  the  middle  frustum  of  a 
prolate  spheroid  E  F  G  H,  the  mid- 
dle diameter,  B  D,  is  50  inches,  and 
that  of  either  of  the  ends  E  F  or 
G  H,  is  40  inches,  and  its  length, 
WW,  18  inches  ;  what  is  its  solidity? 

Here  [(50=x2+40')xl8]  x  .2618  =  [(2500x2+1600) 
Xl8]x.2618=  [(500a+1600)xl8]x.26l8=(6000  x  18) 
X  .2618  =  118800  X  .2618  =  31101.84  cubic  inches,  the 
solidity  required. 

2.  What  is  the  solidity  of  the  middle  frustum  of  a  prolate 
spheroid,  the  middle  diameter  being  5  feet,  that  of  either  of 
the  two  ends  3  feet,  and  the  distance  of  the  ends  6  feet  8 
inches »  Ans.   102.9746  feet. 


MENSURATION    OF    SOLIDS.  161 

a.  \Yhat  is  the  solidity  of  the  middle  frustum  of  an  oblate 
spheroid,  the  middle  diameter  being  100  inches,  that  of 
either  of  the  ends  SO  inches,  and  the  distance  of  the  ends 
30  inches  ?  ^ns.  248814.72  inches. 

4.  What  is  the  solidity  of  the  middle  frustum  of  a  prolate 
Rpheroid,  the  middle  diameter  being  5  feet,  that  of  either  of 
the  ends  3  feet,  and  the  distance  of  the  ends  0  feet  ? 

^ns.  92.0772  ft. 

Case  2. — When  the  ends  are  elliptical  or  perpendicular 
to  the  revolving  axis. 

Rule  1. — Multiply  twice  the  transverse  diameter  of  the 
middle  section  by  its  conjugate  diameter,  and  to  this  product 
add  the  product  of  the  transverse  and  conjugate  diameters 
of  either  of  the  ends. 

2.  Multiply  the  sum  thus  found,  by  the  distance  of  the 
ends  or  the  height  of  the  frustum,  and  the  product  again  by 
.2018,  and  it  will  give  the  solidity  required. 

EXAMPLES. 

_G 

1.  In  the  middle  frustum,  A  BC  D,       jS^^^Ki^^ 
of  a  prolate  spheroid,  the  diameters  of     iF^^^S^^^B^ 

the  middle  section  are  50  and  30  inches;  ek^ tB^^^B^ 

those  of  the  end  40  and  24  inches ;  and  ^iliSipjSii^B^r 
its  height,  2  on,  18  inches  ;  what  is  the  ^I^SwfmKlS^ 
solidity  ?  "h" 

Here  {[(50x2  x  30)  +  (40x24)]x  18}  x  .2618=[(3000 
+  000)  X  18]  X  .2018  =  (3000  x  18)  x  .2018  =  71280  X 
.2018  =  18001.104  cubic  inches,  the  solidity  required. 

2.  In  the  middle  frustum  of  a  prolate  spheroid,  the  diame- 
ters of  the  middle  section  are  100  and  (50  inches ;  those  of 
the  end  80  and  48  inches ;  and  the  length  30  inches :  what 
is  the  solidity  ?  ^ns.  80.3040  ft. 

3.  In  the  middle  frustum  of  an  oblate  spheroid,  the 
diameters  of  the  middle  section  are  100  and  00  inches; 
those  of  the  end  00  and  30  inches ;  and  the  length  80 
inches  ;  what  is  the  solidity  of  the  frustum  ? 

^ns.   171.0244  a. 


14* 


A  TABLE  OF  THE  AREAS  OF  THE  SEGMENTS 
OF  A  CIRCLE, 

Whose  diameter  is  Unity,  and  supposed  to  be  divided  into 
1000  equal  Parts. 


■  Versed 
1  Sine. 

Seg.  Area. 

Versed 
Sine. 

Seg.  Area. 

Versed 
8iiie. 

1 

Seg.  Area.  1 

.001 

.000042 

.034 

.008273 

.067 

.022(552 

1  .002 

.000119 

.035 

.008638 

.068 

.02;n.54  ' 

.003' 

.000219 

.036 

.009008 

.069 

.023659 

.004 

.000337 

.037 

.009383 

.070 

.024168 

.005 

.000470 

.038 

.009763 

.071 

.024680 

.006 

.000618 

.039 

.010148 

.072 

.025195 

.007 

.000779 

.040 

.010537 

.073 

.025714 

.008 

.000951 

.041 

.010931 

.074 

.026236 

.009 

.001135 

.042 

.011330 

.075 

.026761 

.010 

.001329 

.043 

.011734 

.076 

.027289 

.011 

.001533 

.044 

.012142 

.077 

.027821 

.012 

.001746 

.045 

.012554 

.078 

.028356 

.013 

.001968 

.046 

.012971 

.079 

.028894 

.014 

.002199 

.047 

.013392 

.080 

.029435 

.015 

.002438 

.048 

.013818 

.081 

.029979 

.016 

.002685 

.049 

.014247 

.082 

.030526 

.017 

.002940 

.050 

.014681 

.083 

.031076 

.018 

.003202 

.051 

.015119 

.084 

.031629 

.019 

.003471 

.052 

.015561 

.085 

.032186 

.020 

.003748 

.053 

.016007 

.086 

.032745 

.021 

.004031 

I  .051 

.016457 

.087 

.033307 

.022 

.004322 

I  .055 

.016911 

.088 

.033872 

.023 

.004618 

.056 

.017369 

.089 

.034441 

.024 

.004921 

.057 

.017831 

.090 

.035011  1 

.025 

.005230 

.058 

.018296 

.091 

.035585  j 

.026 

.005546 

.059 

.018766 

.092 

.036162 

.027 

.005867 

.060 

.019239 

.093 

.036741  I 

.038 

.006194 

.061 

.019716 

.094 

.037323 

.029 

.006527 

.062 

.020196 

.095 

.037909 

.030 

.006865 

.063 

.020680 

.096 

.038496 

.031 

.007209 

.064 

.021168 

.097 

.039087 

.032 

.007558 

.065 

.0216.59 

.098 

.039680 

1  .0:33 

1 

.007913 

.086 

.022154 

.099 

.040276 

29.'i 


£96 


THE  AREAS  OF  THE  SEGMENTS  OF  A  CIRCLE. 


Versed 
Sine. 

Seg.  Area. 

Versed 

Sine. 

Seg.  Area.  1 

Versed 

Sine. 

■  -  -1 

Seg.  Ai  (a. 

.100 

.040875 

.141 

.067528 

.182 

.097674 

.101 

.041476 

.142 

.0()8225 

.183 

.098447 

.102 

.042080 

.143 

.068924 

.184 

.099221 

.103 

.042687 

.144 

.069625 

.185 

.099997 

.104 

.043296 

.145 

.070328 

.186 

.100774 

.105 

.043908 

.146 

.071033 

.187 

.101653 

.106 

.044522 

.147 

.071741 

.188 

.102334 

.107 

.045139 

.148 

.072450 

.189 

.103116 

.108 

.045759 

.149 

.073161 

.190 

.103900 

.109 

.046381 

.160 

.073874 

.191 

.10^1685 

.110 

.047005 

.151 

.074589 

.192 

.105472 

.111 

.047632 

.152 

.075306 

.193 

.106261 

.112 

.048262 

.153 

.076026 

.194 

.107051 

.113 

.048894 

.154 

.076747 

.195 

.107842 

.114 

.049528 

.155 

.077469 

.196 

.108636 

.115 

.050165 

.166 

.078194 

.197 

.109430 

.116 

.050804 

.157 

.078921 

.198 

.110226 

.117 

.051446 

.158 

.079649 

.199 

.111024 

.118 

.052090 

.1.59 

.080380 

.200 

.111823 

.119 

.052736 

.160 

.081112 

.201 

.112624 

.120 

.053385 

.161 

.081846 

.202 

.113426 

.121 

.054036 

.162 

.082582 

.203 

.1142530 

.122 

.054689 

.163 

.083320 

.204 

.115035 

.123 

.055345 

.164 

.084059 

.205 

.115842 

.124 

.056003 

.165 

.084801 

.206 

.116650 

.125 

.056663 

.166 

.085544 

.207 

.117460 

.126 

.057326 

.167 

.086289 

.208 

.118271 

.127 

.057991 

.168 

.087036 

.209 

.119083 

.128 

.0.58658 

.169 

.087785 

.210 

.119897 

.129 

.059327 

.170 

.088535 

.211 

.120712 

.130 

.059999 

.171 

.089287 

.212 

.121529 

.131 

.060872 

.172 

.090041 

.213 

.1225347 

.132 

.061348 

.173 

.090797 

.214 

.123167 

.133 

.062026 

.174 

.091.554 

.215 

.123988 

.134 

1  .062707 

.175 

.092313 

.216 

.124810 

.135 

.063389 

.176 

.093074 

.217 

.12.5634 

.136 

.064074 

.177 

.093836 

.218 

.126459 

.137 

.064760 

.178 

.094601 

.219 

.127285 

AiiS 

1  .005449 

.179 

.095366 

.220 

.128113 

.139 

1  .066140 

.180 

.096134 

.221 
.222 

.128942 

■  .140 

j  .0668*3 

.181 

.096903 

.129773 

THE  AREAS  OF  THE  SEGMENTS  OF  A  CIRCLE.        297 


Versed 
Sine. 

Scg.  Area. 

Versed 
Sine. 

Seg.  Area. 

Versed 

Sine. 

Seg.  Area. 

.223 

.130005 

.264 

.165780 

.305 

.202761 

.224 

.131438 

.265 

.166663 

.306 

.20:3683 

.225 

.132272 

.266 

.167546 

.307 

.204605 

.226 

.133108 

.267 

.168430 

.308 

.205527 

.227 

.1.33945 

.268 

.169315 

.309 

.206451 

.228 

.134784 

.260 

.170202 

.310 

.207:376 

.229 

.135624 

.270 

.171089 

.311 

.208301 

.230 

.136465 

.271 

.171978 

.312 

.209227 

.231 

.137307 

.272 

.172867 

.313 

.210154 

.232 

.138150 

.273 

.173758 

.314 

.211082 

.2:« 

.138995 

.274 

.174049 

.315 

.212011 

.2:34 

.139841 

.275 

.175542 

.316 

.212940" 

.235 

.140(J88 

.276 

.176435 

.317 

.213871 

.236 

.141537 

.277 

.177330 

.318 

.214802 

.2:37 

.142387 

.278 

.178225 

.319 

.215733 

.238 

•1432:38 

.279 

.179122 

.320 

.216666 

.239 

.144091 

.280 

.180019 

.321 

.217599 

.240 

.144944 

.281 

.180918 

.322 

.218533 

.241 

.145799 

.282 

.181817 

.323 

.219468 

.242 

.146655 

.283 

.182718 

.324 

.220404 

.243 

.147512 

.284 

.183619 

.325 

.221340 

.244 

.148371 

.285 

.184521 

.32() 

.222277 

.245 

.149230 

.286 

.185425 

.327 

.22:3215 

.246 

.1500!)  1 

.287 

.186329 

.328 

.224154  ! 

.247 

.150953 

.288 

.187234 

.329 

.225093 

.248 

.151816 

.289 

.188140 

.330 

.22(5033 

.249 

.152680 

.290 

.189047 

.331 

.226974 

.250 

.153546 

.291 

.189955 

.332 

.227915 

.251 

.15^1412 

.292 

.190864 

.333 

.228858  ' 

.252 

.155280 

.293 

.191775 

.3:34 

.229801 

.253 

.156149 

.294 

.192684 

.335 

.230745 

.254 

.157019 

.295 

.193596 

.336 

.231689 

.255 

.157890 

.296 

.194509 

.337 

.2326:34 

.256 

.158762 

.297 

.195422 

.338 

.233.580 

.257 

.159636 

.298 

.196337 

.339 

.234526 

.258 

.160510 

.299 

.197252 

.340 

.235473 

.259 

.161386  ' 

.300 

.198168 

.:J41 

.236421 

.260 

.162263 

.301 

.199085 

.342 

.237369 

.261 

.163140  i 

.302 

.200003 

.343 

.238318 

.262 

.164019 

.303 

.200922 

.344 

.239268 

.263 

.164899 

.304 

.201841 

.345 

.240218  1 

298 


THE  AREAS  OF  THE  SEGMENTS  OF  A  CIBCLE. 


Versed 
Sine. 

Scg.  Area. 

Versed 
Sine. 

Seg.  Area. 

Versed 
Sine. 

Seg.  Area. 

.346 

.5^1169 

.387 

.280668 

.428 

.320948 

.347 

.242121 

.388 

.281642 

.429 

.321938 

.348 

.243074 

.389 

.282617 

.430 

.322928 

.349 

.244026 

.390 

.283592 

.431 

.323918 

.350 

.244980 

.391 

.284568 

.432 

.324909 

.351 

.^5934 

.392 

.285544 

.433 

.325900 

.352 

.246889 

.393 

.286521 

.434 

.326892 

.353 

.247845 

.394 

.287498 

.435 

.327882 

.354 

.248801 

.395 

.288476 

.436 

.328874 

.355 

.249757 

.396 

.289453 

.437 

.329866 

.356 

.250715 

.397 

.290432 

.438 

.330858 

.357 

.251673 

.398 

.291411 

.439 

.331850 

.358 

.252631 

.399 

.292390 

.440 

.332843 

.359 

.253590 

.400 

.293369 

.441 

.833836 

.360 

.254550 

•401 

.294349 

.442 

.334829 

.361 

.255510 

.402 

.295330 

.443 

.335822  1 

.362 

.256471 

.403 

.296311 

.444 

.336816 

.363 

.257433 

.404 

.297292 

.445 

.337810 

.364 

.258395 

.405 

.298273 

.446 

.338804 

.365 

.259357 

.406 

.299255 

.447 

.339798 

.366 

.260320 

.407 

.300288 

.448 

.340793 

.367 

.261284 

•408 

.301220 

.449 

.341787 

.368 

.262248 

.409 

.302203 

.450 

.342782 

.369 

.263213 

.410 

.303187 

.451 

.343777 

.370 

.264178 

.411 

.304171 

.452 

.344772 

.371 

.265144 

.412 

.305155 

.453 

.845768 

.372 

.266111 

.413 

.306140 

.454 

.346764 

.373 

.267078 

.414 

.307125 

.455 

.347759 

.374 

.268045 

.415 

.308110 

.456 

.348755 

.375 

.269013 

.416 

.309095 

.457 

.349752 

.376 

.269982 

.417 

.310081 

.458 

.350748 

.377 

.270951 

.418 

.311068 

.459 

.851745 

.378 

.271920 

.419 

.312054 

.460 

.352741 

.379 

.272890 

.420 

.318041 

.461 

.353739 

.380 

.273861 

.421 

.314029 

.462 

.354736 

.381 

.274882 

.422 

.315016 

.463 

.355782 

382 

.275803 

.423 

.316004 

.464 

.856780 

.383 

.276775 

.424 

.316992 

.465 

.357727  1 

.384 

.277748 

.425 

.317981 

.466 

.358725 

.385 

.278721 

.426 

.318970 

.467 

.359723 

.386 

.279694  j 

.427 

.319959 

.468 

.360721  i 

THE  AREAS  OF  THE  SEGMENTS  OF  A  CIRCLE. 


299 


Versed 
Sine. 

Seg.  Area. 

Versed 
Sine. 

Seg.  Area. 

Versed 
Sine. 

Seg.  J^tea. 

.469 

.361719 

.480 

.372764 

.491 

.383699 

.470 

.362717 

.481 

.373703 

.492 

384699 

.471 

.363715 

.482 

.374702 

.493 

.385699 

.472 

.364713 

.483 

.375702 

.494 

.386699 

.473 

.365712 

.484 

.376702 

.495 

.387699 

.474 

.366710 

.485 

.377701 

.496 

.388699 

.475 

.367709 

.486 

.378701 

.497 

.389699 

.476 

.368708 

.487 

.379700 

.498 

.390699 

.477 

.369707 

.488 

.380700 

.499 

.391699 

.478 

.370706 

.489 

.381699 

.500 

.39269y 

.479 

.371705 

.490 

.382699 

THE  END. 


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